CAT 2025 Quant Questions — Slot 3

Picture the number first. $10^{50}+10^{25}$ is just two $1$'s sitting far apart in a sea of zeros; subtracting $123$ only disturbs the bottom end, forcing a chain of borrows. Build the digits, then add them.

Step 1 — What is $10^{50}+10^{25}$? A $1$ in the $50$th place and a $1$ in the $25$th place:

$1\underbrace{00\ldots0}_{24\text{ zeros}}1\underbrace{00\ldots0}_{25\text{ zeros}}.$

Step 2 — Subtract $123$. The subtraction only touches the lowest $25$-digit block, which is exactly $10^{25}$. So compute $10^{25}-123$:

$10^{25}-123=\underbrace{99\ldots9}_{22\text{ nines}}877.$

(The last three digits come from $1000-123=877$; each of the $22$ digits above them drops to $9$ through the borrow.)

Step 3 — Add up all the digits.

• The leading $1$ (at place $50$): contributes $1$.
• The $24$ middle zeros: contribute $0$.
• The bottom block $\underbrace{9\ldots9}_{22}877$: contributes $22\times 9+8+7+7=198+22=220$.

$\text{Digit sum}=1+0+220=221.$

Answer: $\boxed{221}$

💡 Takeaway: for powers-of-ten minus a small number, never multiply anything out — track only the borrow chain, since that's the sole place the digits actually change.

Two facts unlock this: a quadrilateral with an inscribed circle has its pairs of opposite sides summing equally, and its inradius equals area divided by semi-perimeter. Set coordinates and turn both facts into equations.

$ABCD$ is a right trapezium: $AB\parallel DC$, $AD\perp AB$, and $AB=3\,DC$.

Step 1 — Coordinates. Place $A=(0,0)$, $B=(3x,0)$, $D=(0,h)$, $C=(x,h)$, so $DC=x$ and $AB=3x$ (matching $AB=3\,DC$). The slant side is

$BC=\sqrt{(3x-x)^2+h^2}=\sqrt{4x^2+h^2}.$

Step 2 — The tangential-quadrilateral condition $AB+DC=AD+BC$ (opposite sides sum equally):

$3x+x=h+\sqrt{4x^2+h^2}\ \Longrightarrow\ \sqrt{4x^2+h^2}=4x-h.$

Square both sides:

$4x^2+h^2=16x^2-8xh+h^2\ \Longrightarrow\ 8xh=12x^2\ \Longrightarrow\ h=\frac{3x}{2}.$

Step 3 — Area, perimeter, inradius.

$\text{Area}=\tfrac12(AB+DC)\,h=\tfrac12(4x)\left(\tfrac{3x}{2}\right)=3x^2.$

$BC=\sqrt{4x^2+\tfrac{9x^2}{4}}=\tfrac{5x}{2},\qquad \text{perimeter}=3x+x+\tfrac{3x}{2}+\tfrac{5x}{2}=8x,$

so the semi-perimeter is $s=4x$ and the inradius is

$r=\frac{\text{Area}}{s}=\frac{3x^2}{4x}=\frac{3x}{4}.$

Step 4 — Use $r=3$. $\dfrac{3x}{4}=3\Rightarrow x=4$, hence

$\text{Area}=3x^2=3\times 16=48\text{ cm}^2.$

Answer: $\boxed{48\text{ cm}^2}$

💡 Teacher tip: the moment a problem says "a circle is inscribed touching all sides", reach for opposite-sides-sum-equally and $r=\text{Area}/s$ — together they usually crack the figure.

The hidden circle: the condition $\angle ACB=\angle AEB$ means $C$ and $E$ see the segment $AB$ at the same angle, so $A,B,C,E$ all lie on one circle. Then the intersecting-chords rule finishes it.

Here $AB=AC=12$, $D$ lies on $BC$ with $AD=8$, and $AD$ is extended to $E$ with $\angle ACB=\angle AEB$.

Step 1 — Why the four points are concyclic. Angles $\angle ACB$ and $\angle AEB$ both stand on the same chord $AB$ from the same side. Equal angles on the same chord means $A,B,C,E$ lie on a common circle.

Step 2 — A neat length fact in the isosceles triangle. Drop the altitude $AM$ from $A$ to $BC$ ($M$ the midpoint). For any point $D$ on $BC$,

$BD\cdot DC=BM^2-MD^2.$

Also $BM^2+AM^2=AB^2$ and $AM^2+MD^2=AD^2$, so

$BD\cdot DC=AB^2-AM^2-MD^2=AB^2-AD^2=12^2-8^2=144-64=80.$

Step 3 — Intersecting chords at $D$. Chords $AE$ and $BC$ cross at $D$, so

$AD\cdot DE=BD\cdot DC=80.$

With $AD=8$: $DE=\dfrac{80}{8}=10$.

Step 4 — Length $AE$.

$AE=AD+DE=8+10=18\text{ cm}.$

Answer: $\boxed{18\text{ cm}}$

💡 Why it's slick: you never need $BC$ or the exact position of $D$. The identity $BD\cdot DC=AB^2-AD^2$ uses only the two given lengths, and the chord theorem hands you $DE$ at once.

Strategy: track the total salary, not the averages. A percentage change in the overall average is really a change in the total pay, and here that change comes only from the managers.

There are $5$ managers and $25$ engineers — $30$ people — averaging $60000$.

Step 1 — Total pay before the raise.

$T=30\times 60000=1{,}800{,}000.$

Step 2 — Total pay after a 5% higher average. The new average is $60000\times 1.05=63000$, so

$T'=30\times 63000=1{,}890{,}000.$

Step 3 — Whose money changed? Only the managers got a raise, so the extra $T'-T=90{,}000$ is entirely their $20\%$ increase. If $M$ is the managers' total salary,

$0.20\,M=90{,}000\ \Longrightarrow\ M=450{,}000.$

Step 4 — Engineers' average. The engineers' total is $T-M=1{,}800{,}000-450{,}000=1{,}350{,}000$, shared by $25$ engineers:

$\frac{1{,}350{,}000}{25}=54{,}000.$

Answer: $\boxed{54000}$

💡 Takeaway: convert every "average" into a "total" early. Averages don't add cleanly, but totals do — and the raise then attaches to exactly one group.

Think of it as: one unknown marked price runs through four months of different discounts. Write each month's profit in terms of that price, add them, and match the given total.

Step 1 — Profit per unit at marked price $M$. Cost is ₹240, so after a discount $d$ the profit per unit is $M(1 - d) - 240$.

Step 2 — Add up and set equal to ₹1,38,825. The $M$-coefficients sum to $96 + 121.5 + 142.5 + 165 = 525$; the constants sum to $136800$: $525M - 136800 = 138825 \;\Rightarrow\; 525M = 275625 \;\Rightarrow\; M = \frac{275625}{525} = 525.$

Answer: $\boxed{525}$ (Choice B)

💡 Teacher tip: with one fixed marked price across several discount periods, keep everything symbolic in $M$ and collect the coefficient once. The four separate profit lines then reduce to a single linear equation.

Think of it as: an arithmetic progression is fully described by its first term $a$ and common difference $d$. Two conditions give two equations — solve, then answer.

Step 1 — Translate both conditions using $a_n = a + (n-1)d$.

The fourth, seventh and tenth terms sum to $99$: $(a + 3d) + (a + 6d) + (a + 9d) = 3a + 18d = 99 \;\Rightarrow\; a + 6d = 33. \qquad (1)$

The first fourteen terms sum to $497$: $S_{14} = \frac{14}{2}(2a + 13d) = 7(2a + 13d) = 497 \;\Rightarrow\; 2a + 13d = 71. \qquad (2)$

Step 2 — Solve. From $(1)$, $a = 33 - 6d$. Substitute into $(2)$: $2(33 - 6d) + 13d = 71 \;\Rightarrow\; 66 + d = 71 \;\Rightarrow\; d = 5, \quad a = 3.$

Step 3 — Sum of the first five terms. $S_5 = \frac{5}{2}(2a + 4d) = \frac{5}{2}(6 + 20) = \frac{5}{2} \times 26 = 65.$

Answer: $\boxed{65}$

💡 Teacher tip: the middle term of three equally spaced AP terms equals their average, so $a_4 + a_7 + a_{10} = 3a_7$. That instantly gives $a_7 = 33$ — a handy shortcut worth spotting.

Think of it as: everything hinges on one person's daily output. Find each team's per-member rate, track the job to the moment team C's member leaves, then see how many team-B members cover the shortfall.

Step 1 — Per-member rates. Each member of a team works equally, so divide the team rate by the team size.

Step 2 — Rate of the starting crew (2 from A, 3 from B, 1 from C): $\frac{2}{200} + \frac{3}{400} + \frac{1}{40} = \frac{4}{400} + \frac{3}{400} + \frac{10}{400} = \frac{17}{400} \text{ per hour}.$

Step 3 — Work done in the first $23$ hours. $23 \times \frac{17}{400} = \frac{391}{400}, \quad\text{so remaining} = 1 - \frac{391}{400} = \frac{9}{400}.$

Step 4 — Finish the last bit in one hour. After C's member leaves, the crew (2A + 3B) works at $\tfrac{4}{400} + \tfrac{3}{400} = \tfrac{7}{400}$ per hour. To clear $\tfrac{9}{400}$ in one hour we need an extra $\tfrac{9}{400} - \tfrac{7}{400} = \tfrac{2}{400}$ per hour. Each added B member supplies $\tfrac{1}{400}$, so we need $\frac{2/400}{1/400} = 2 \text{ members}.$

Answer: $\boxed{2}$ (Choice B)

💡 Teacher tip: in mixed-crew work problems, always drop down to the per-person rate first. Once you know what one member contributes, "how many more do we need" becomes a one-step division.

The standard tool: to find the range of a rational function, set $y=f(x)$, clear the denominator into a quadratic in $x$, and demand that this quadratic has a real solution (discriminant $\ge 0$). The allowed $y$-values are exactly the range.

Step 1 — Set $y=f(x)$ and clear the denominator.

$y=\frac{2x-3}{2x^2+4x-6}\ \Longrightarrow\ y(2x^2+4x-6)=2x-3,$

which rearranges to the quadratic in $x$:

$2y\,x^2+(4y-2)x+(3-6y)=0.$

Step 2 — Require a real $x$. For a real $x$ to exist (with $y\ne 0$; the value $y=0$ is reached at $x=\tfrac32$), the discriminant must be $\ge 0$:

$(4y-2)^2-4(2y)(3-6y)\ge 0.$

Step 3 — Expand and simplify.

$16y^2-16y+4-24y+48y^2\ge 0\ \Longrightarrow\ 64y^2-40y+4\ge 0.$

Divide by $4$:

$16y^2-10y+1\ge 0.$

Step 4 — Solve the quadratic inequality. Its roots:

$y=\frac{10\pm\sqrt{100-64}}{32}=\frac{10\pm 6}{32}\ \Longrightarrow\ y=\frac12\ \text{or}\ y=\frac18.$

Since the parabola $16y^2-10y+1$ opens upward, it is $\ge 0$ outside the roots:

$y\le \frac18\quad\text{or}\quad y\ge \frac12.$

Answer: $\boxed{\left(-\infty,\tfrac18\right]\cup\left[\tfrac12,\infty\right)}$

💡 Teacher tip: the "discriminant $\ge 0$" method returns the range of any rational function whose top is lower-degree than its bottom. Just check separately any $y$ that makes the leading coefficient ($2y$ here) vanish.

Set-up: a three-subject Venn diagram with one relationship tying the overlaps together. Use the total-students identity to pin down the "all three" region, then push the target region to its maximum.

There are $150$ students; Physics $75$, Maths $111$, Chemistry $40$. Everyone takes at least one subject.

Step 1 — Name the overlaps using the given relation. Let the Physics–Maths overlap be $2b$. We're told Physics–Chemistry $=$ Chemistry–Maths $=$ half of that $=b$. Let $t$ be the number taking all three.

Step 2 — Apply inclusion–exclusion (union $=150$).

$150=75+111+40-(2b+b+b)+t=226-4b+t,$

so $t=4b-76$.

Step 3 — Write each region. Let $p,m,c$ be the "only one subject" counts. Standard bookkeeping gives, after substituting $t=4b-76$:

$p=b-1,\qquad m=b+35,\qquad c=2b-36.$

Step 4 — Bound $b$ using non-negativity.

• The "exactly Physics–Chemistry" region $=b-t=76-3b\ge 0\Rightarrow b\le 25$.
• At least one student takes all three: $t=4b-76\ge 1\Rightarrow b\ge 20$.

So $b$ ranges over $20$ to $25$.

Step 5 — The target: Physics but not Maths. That's all of Physics minus its overlap with Maths, $=p+(b-t)$:

$\#(P\setminus M)=(b-1)+(76-3b)=75-2b.$

This is largest when $b$ is smallest, i.e. $b=20$:

$75-2(20)=35.$

Step 6 — Verify at $b=20$, $t=4$. Only-regions $p=19,\ m=55,\ c=4$; pairwise-only regions $36,16,16$; all three $=4$. Total $=19+55+4+36+16+16+4=150$ ✓, and the overlaps are $40,20,20$ as required ✓.

Answer: $\boxed{35}$

💡 Read the target carefully: "Physics but not Maths" is the whole Physics circle outside Maths — the Physics-only slice plus the Physics-and-Chemistry-but-not-Maths slice — not just "Physics only".

Plan: every base is built from the primes $2$, $3$, $5$. Rewrite both sides as $2^{?}\,3^{?}\,5^{?}$ and match the exponents of each prime separately.

The equation is $12^{12x}\times 4^{24x+12}\times 5^{2y}=8^{4z}\times 20^{12x}\times 243^{3x-6}$.

Step 1 — Factor the bases. $12=2^2\cdot 3,\ 4=2^2,\ 8=2^3,\ 20=2^2\cdot 5,\ 243=3^5$.

Step 2 — Left side as prime powers.

$(2^2\cdot 3)^{12x}(2^2)^{24x+12}\,5^{2y}=2^{24x+48x+24}\,3^{12x}\,5^{2y}=2^{72x+24}\,3^{12x}\,5^{2y}.$

Step 3 — Right side as prime powers.

$(2^3)^{4z}(2^2\cdot 5)^{12x}(3^5)^{3x-6}=2^{12z+24x}\,3^{15x-30}\,5^{12x}.$

Step 4 — Match each prime.

Step 5 — Add.

$x+y+z=10+60+42=112.$

Answer: $\boxed{112}$

💡 Takeaway: whenever an equation multiplies many bases together, the fastest route is almost always to break everything into common primes and equate exponents — one clean equation per prime.

Plan: eliminate $q$ so the constraint $0.3q\le p\le 0.7q$ becomes a window for $p$ in terms of $r$. Then sweep $r$ over its legal perfect squares to find $p$'s largest and smallest possible values.

We have natural numbers with $p+q+r=900$; $r$ is a perfect square with $150<r<500$; and $0.3q\le p\le 0.7q$.

Step 1 — Remove $q$. Since $q=900-r-p$, the two bounds become

$0.3(900-r-p)\le p\le 0.7(900-r-p).$

Step 2 — Solve each side for $p$.

• Upper: $p\le 0.7q\Rightarrow 1.7p\le 0.7(900-r)\Rightarrow p\le \dfrac{6300-7r}{17}$.
• Lower: $p\ge 0.3q\Rightarrow 1.3p\ge 0.3(900-r)\Rightarrow p\ge \dfrac{2700-3r}{13}$.

Step 3 — See how each bound moves with $r$. Both $\dfrac{6300-7r}{17}$ and $\dfrac{2700-3r}{13}$ decrease as $r$ grows. So the largest $p$ comes from the smallest allowed $r$, and the smallest $p$ from the largest allowed $r$.

The perfect squares strictly between $150$ and $500$ run from $13^2=169$ up to $22^2=484$. Smallest is $169$, largest is $484$.

Step 4 — Maximum $p$ (at $r=169$).

$p\le \frac{6300-7(169)}{17}=\frac{6300-1183}{17}=\frac{5117}{17}=301.$

Check: $q=900-169-301=430$, and $0.3(430)=129\le 301\le 301=0.7(430)$ ✓.

Step 5 — Minimum $p$ (at $r=484$).

$p\ge \frac{2700-3(484)}{13}=\frac{2700-1452}{13}=\frac{1248}{13}=96.$

Check: $q=900-484-96=320$, and $0.3(320)=96\le 96\le 224=0.7(320)$ ✓.

Step 6 — Add the extremes.

$301+96=397.$

Answer: $\boxed{397}$

💡 Takeaway: when a target variable is squeezed by an interval that depends on another quantity, check whether the bounds rise or fall with that quantity — the extremes then sit at the ends of its allowed range.

Spot the structure: the two brackets $x^2+3x$ and $x^2+3x+2$ share the block $x^2+3x$. Naming that block turns a messy quartic into something you can complete into a perfect square.

Step 1 — Substitute. Let $u=x^2+3x$. Then

$f(x)+1=u(u+2)+1=u^2+2u+1=(u+1)^2.$

So the equation $f(x)+1=9701$ becomes

$(u+1)^2=9701\ \Longrightarrow\ u+1=\pm\sqrt{9701}\ \Longrightarrow\ u=-1\pm\sqrt{9701}.$

Step 2 — Which values of $u$ give real $x$? Recall $u=x^2+3x$, i.e. $x^2+3x-u=0$. This has real solutions only when its discriminant is non-negative:

$9+4u\ge 0\ \Longrightarrow\ u\ge -\tfrac94=-2.25.$

Now test the two candidates ($\sqrt{9701}\approx 98.5$):

Step 3 — Sum the real roots. Only the first $u$ produces real values of $x$, from $x^2+3x-u=0$. By the sum-of-roots rule (sum $=-\,$coefficient of $x$), those two roots add to $-3$.

Answer: $\boxed{-3}$

💡 Why it's elegant: the roots of $x^2+3x-u=0$ always add to $-3$, whatever $u$ is. So the ugly $\sqrt{9701}$ never touches the final answer — you only needed to confirm which branch of $u$ is real.

Think of it as: four unknown digits tied together by three simple relationships. Turn those relationships into equations, express everything in terms of one digit, then list the numbers the digit rules actually allow.

Let the number be $\overline{abcd}$, where $a$ is the thousands digit, $b$ hundreds, $c$ tens, $d$ units. Since it's a genuine 4-digit number, $a$ runs from $1$ to $9$ and $b,c,d$ from $0$ to $9$.

Step 1 — Write the three conditions.

$a+b+c=15,\qquad b+c+d=16,\qquad c=d+6.$

Step 2 — Reduce to one digit. Subtract the first equation from the second: the $b+c$ cancels, leaving $d-a=1$, so $a=d-1$. The third gives $c=d+6$. Put $c$ into the second equation:

$b=16-c-d=16-(d+6)-d=10-2d.$

So every digit is now written in terms of $d$: $a=d-1,\ b=10-2d,\ c=d+6$.

Step 3 — Apply the digit limits. Each expression must be a legal digit:

• $c=d+6\le 9\Rightarrow d\le 3$.
• $a=d-1\ge 1\Rightarrow d\ge 2$.
• $b=10-2d\ge 0\Rightarrow d\le 5$ (already satisfied).

So $d$ can only be $2$ or $3$.

Step 4 — List the numbers.

Step 5 — Take the difference. Largest minus smallest:

$2493-1682=811.$

Answer: $\boxed{811}$

💡 Takeaway: when several digit-sum conditions overlap, subtracting one equation from another kills the shared terms and collapses the whole problem to a single free digit.

Two-stage mixing: pour out of A into B, stir, then pour the same volume back. Track how much alcohol lives in A at the end and compare with the final ratio.

Vessel A starts with $60$ L alcohol; B starts with $60$ L water. Let $v$ litres be moved each time.

Step 1 — First pour ($v$ L from A to B). A now holds $60-v$ L of alcohol. B becomes $60+v$ L of liquid containing $v$ L alcohol, so its alcohol fraction is $\dfrac{v}{60+v}$.

Step 2 — Second pour ($v$ L from B back to A). That returning $v$ L carries

$\text{alcohol}=\frac{v^2}{60+v},\qquad \text{water}=\frac{60v}{60+v}.$

So A's alcohol becomes

$(60-v)+\frac{v^2}{60+v}=\frac{(60-v)(60+v)+v^2}{60+v}=\frac{3600-v^2+v^2}{60+v}=\frac{3600}{60+v},$

and A's water is $\dfrac{60v}{60+v}$.

Step 3 — Apply the final ratio $15:4$ in A.

$\frac{\text{alcohol}}{\text{water}}=\frac{3600/(60+v)}{60v/(60+v)}=\frac{3600}{60v}=\frac{60}{v}.$

Set this equal to $\dfrac{15}{4}$:

$\frac{60}{v}=\frac{15}{4}\ \Longrightarrow\ v=\frac{60\times 4}{15}=16.$

Answer: $\boxed{16\text{ litres}}$

💡 Takeaway: notice how A's final alcohol simplified to $\dfrac{3600}{60+v}$ — the neat cancellation $(60-v)(60+v)+v^2=3600$ is what makes the ratio collapse to $\dfrac{60}{v}$.

The key observation: the stop happens at the same point on the road in both scenarios — only how long he rests changes. That lets us write two "total journey = 6 hours" equations and subtract them.

The trip is scheduled from 5 pm to 11 pm, so the planned time is $6$ hours. Let his initial speed be $v$ km/h, so the whole distance is $D=6v$.

Let him drive for time $t$ at speed $v$, then stop, then finish the rest of the distance at a higher speed.

Step 1 — The two scenarios.

Here $D-vt$ is the distance still left when he stops (same in both cases).

Step 2 — Subtract scenario 1 from scenario 2. The $t$ terms cancel:

$\frac16+(D-vt)\left(\frac{1}{v+5}-\frac{1}{v+3}\right)=0.$

Combining the fractions, $\dfrac{1}{v+5}-\dfrac{1}{v+3}=\dfrac{-2}{(v+3)(v+5)}$, so

$D-vt=\frac{(v+3)(v+5)}{12}.$

Step 3 — Feed this back into scenario 1.

$t+\frac13+\frac{(v+3)(v+5)/12}{v+3}=6\ \Longrightarrow\ t+\frac13+\frac{v+5}{12}=6,$

which gives $t=\dfrac{63-v}{12}$.

Step 4 — Use the total distance $D=6v$. Since $D-vt=\dfrac{(v+3)(v+5)}{12}$:

$6v-v\cdot\frac{63-v}{12}=\frac{(v+3)(v+5)}{12}.$

Multiply through by $12$:

$72v-v(63-v)=(v+3)(v+5)\ \Longrightarrow\ 72v-63v+v^2=v^2+8v+15.$

The $v^2$ cancels: $9v=8v+15$, so $v=15$.

Step 5 — Verify. With $v=15$: $D=90$ km, $t=4$ h, so he covers $60$ km before stopping and has $30$ km left.

• Scenario 1: $4+\tfrac13+\tfrac{30}{18}=4+0.333+1.667=6$ h. ✓
• Scenario 2: $4+\tfrac12+\tfrac{30}{20}=4+0.5+1.5=6$ h. ✓

Answer: $\boxed{15\text{ km/h}}$

💡 Takeaway: two "same total time" situations differing in one small way are made for subtraction — the common unknowns vanish and you're left with a clean relationship.

Plan: use the two given ratios to find the actual coin counts, then a $1:1$ split is just "send each box to half the grand total".

Step 1 — Set up the initial counts. The ratio of A to B is $17:7$, so write A $=17k$ and B $=7k$.

Step 2 — Use the second ratio. After moving $108$ coins from A to B, the ratio is $37:20$:

$\frac{17k-108}{7k+108}=\frac{37}{20}.$

Cross-multiply:

$20(17k-108)=37(7k+108)\ \Longrightarrow\ 340k-2160=259k+3996.$

$81k=6156\ \Longrightarrow\ k=76.$

So initially A $=17(76)=1292$ and B $=7(76)=532$; the grand total is $1824$.

Step 3 — Check the first shift. A $=1292-108=1184$, B $=532+108=640$, and $1184:640=37:20$ ✓.

Step 4 — Reach a $1:1$ split. Equal boxes means each holds $\dfrac{1824}{2}=912$. A currently has $1184$, so it must give away

$1184-912=272\text{ coins}.$

Answer: $\boxed{272}$

💡 Takeaway: the coin total never changes, so a $1:1$ target is simply "half the total in each box" — no need for fresh variables once you know the counts.

Two-part job: first note the hidden domain restrictions a logarithm forces, then combine the logs into one equation — and only keep the roots that survive the domain.

The equation is $\log_{x-3}(x^2-9)=\log_{x-3}(x+1)+2$.

Step 1 — Domain conditions (write these before solving). For $\log_{x-3}$ to make sense, the base must be positive and not $1$: $x-3>0$ and $x-3\ne 1$, i.e. $x>3$ and $x\ne 4$. The arguments $x^2-9$ and $x+1$ are then automatically positive.

Step 2 — Turn the "$+2$" into a log. Since $2=\log_{x-3}(x-3)^2$,

$\log_{x-3}(x+1)+2=\log_{x-3}\big((x+1)(x-3)^2\big).$

Equating the arguments:

$x^2-9=(x+1)(x-3)^2.$

Step 3 — Simplify. Note $x^2-9=(x-3)(x+3)$. Since $x>3$ we can divide both sides by $(x-3)$:

$x+3=(x+1)(x-3)=x^2-2x-3\ \Longrightarrow\ x^2-3x-6=0.$

Step 4 — Solve and screen against the domain.

$x=\frac{3\pm\sqrt{9+24}}{2}=\frac{3\pm\sqrt{33}}{2}.$

Numerically, $\sqrt{33}\approx 5.74$, giving $x\approx 4.37$ or $x\approx -1.37$.

Step 5 — Sum of valid values. Only one root survives, so the sum is that root itself.

Answer: $\boxed{\dfrac{3+\sqrt{33}}{2}}$

💡 Takeaway: in log equations, list the base and argument restrictions first. Here they quietly delete one of the two algebraic roots, so "sum of roots $=3$" (the tempting trap) is wrong.

Idea: don't try to find $x$ itself. Build the ladder of values $x^k+\tfrac{1}{x^k}$ step by step, because each one links neatly to the ones below it.

We're given $x^2+\dfrac{1}{x^2}=25$ with $x>0$.

Step 1 — Get $x+\dfrac1x$. Since $\left(x+\dfrac1x\right)^2=x^2+\dfrac1{x^2}+2=25+2=27$, and $x>0$ forces the positive root:

$x+\frac1x=\sqrt{27}=3\sqrt3.$

Step 2 — Get $x^3+\dfrac1{x^3}$. Using $\left(x+\tfrac1x\right)^3=x^3+\tfrac1{x^3}+3\left(x+\tfrac1x\right)$:

$x^3+\frac1{x^3}=(3\sqrt3)^3-3(3\sqrt3)=81\sqrt3-9\sqrt3=72\sqrt3.$

Step 3 — Get $x^4+\dfrac1{x^4}$. Square the given value: $\left(x^2+\tfrac1{x^2}\right)^2=x^4+\tfrac1{x^4}+2$, so

$x^4+\frac1{x^4}=25^2-2=623.$

Step 4 — Combine to reach the 7th power. The handy identity is

$\left(x^3+\frac1{x^3}\right)\left(x^4+\frac1{x^4}\right)=x^7+\frac1{x^7}+\left(x+\frac1x\right),$

because the cross terms give $x^{3-4}+x^{4-3}=x+\tfrac1x$. Rearranging:

$x^7+\frac1{x^7}=\left(72\sqrt3\right)(623)-3\sqrt3=\sqrt3\,(72\times 623-3).$

Now $72\times 623=44856$, so $44856-3=44853$:

$x^7+\frac1{x^7}=44853\sqrt3.$

Answer: $\boxed{44853\sqrt3}$

💡 Takeaway: for symmetric powers $x^k+\tfrac1{x^k}$, the rule $\left(x^m+\tfrac1{x^m}\right)\left(x^n+\tfrac1{x^n}\right)=\left(x^{m+n}+\tfrac1{x^{m+n}}\right)+\left(x^{m-n}+\tfrac1{x^{m-n}}\right)$ lets you climb to any power without ever solving for $x$.