Using the logarithmic property that&nbsp;$\log_{a^p}b=\frac{1}{p}\log_ab$$4 \log_{10} x + 4 \log_{100} x + 8 \log_{1000} x = 13$Can be written as&nbsp;$4\log_{10}x+2\log_{10}x+\frac{8}{3}\log_{10}x=13$$\frac{26}{3}\log_{10}x=13$$\log_{10}x=1.5$$x=10^{1.5}$$x=\sqrt{1000}$$\left[\sqrt{1000}\right]=31$Where [.] is Greatest Integer&nbsp;Function since that is what is asked in the question,&nbsp;31 is the greatest integer that does not exceed x.&nbsp;

$(a + b \sqrt{n})$ is the positive square root of $(29 - 12\sqrt{5})$So&nbsp;$29-12\sqrt{5}=\left(a+b\sqrt{n}\right)^2$$29-12\sqrt{5}=a^2+b^2n+2ab\sqrt{n}$$a^2+b^2n=29$ and$ab\sqrt{n}=-6\sqrt{5}$$a^2b^2n=180$$b^2n=\frac{180}{a^2}$Substituting this in the above equation,&nbsp;$a^2+\frac{180}{a^2}=29$$a^4-29a^2+180=0$$a^2=\frac{\left(29\pm\sqrt{29^2-4\left(180\right)}\right)}{2}$$a^2=\frac{\left(29+\sqrt{841-720}\right)}{2}$$a^2=9\ or\ 20$That means, one of&nbsp;$a^2\ or\ b^2n$ is 9 and 20.&nbsp;We also have,&nbsp;$ab\sqrt{n}=-6\sqrt{5}$ that means one of a or b should be negativeAnd also the&nbsp;fact that this is a positive square root,And we need to maximise the value of a, b and n.&nbsp;We can have a=-3, b=1 and n=20.&nbsp;This satisfies all the above equations, and the value of a+b+n=18.&nbsp;

The first term of the expression can be rewritten as&nbsp;$\frac{\frac{1}{3}\log_2\left(a+b\right)}{\log_2c}$Using the property&nbsp;$\frac{m}{n}\log_ab=\log_ab^{\frac{m}{n}}$ this can be rewritten as$\frac{\log_2\left(a+b\right)^{\frac{1}{3}}}{\log_2c}$And finally using the property&nbsp;$\frac{\log_ba}{\log_bc}=\log_ca$, we can rewrite the expression as&nbsp;$\log_c\left(a+b\right)^{\frac{1}{3}}$Doing identical operations in the second term, we get the entire left-hand side to be:$\log_c\left(a+b\right)^{\frac{1}{3}}+\log_c\left(a-b\right)^{\frac{1}{3}}$Using property&nbsp;$\log_ca+\log_cb=\log_c\left(ab\right)$ we get$\log_c\left[\left(a+b\right)^{\frac{1}{3}}\left(a-b\right)^{\frac{1}{3}}\right]$$\log_c\left[\left(a+b\right)\left(a-b\right)\right]^{\frac{1}{3}}$$\log_c\left[\left(a^2-b^2\right)\right]^{\frac{1}{3}}$This expression is given to be equal to 2/3Using the definition of log:&nbsp;$\log_cN=a\ $ which is&nbsp;$c^a=N$we get:$c^{\frac{2}{3}}=\left(a^2-b^2\right)^{\frac{1}{3}}$Cubing both sides:$c^2=a^2-b^2$Finally giving&nbsp;$a^2=b^2+c^2$We have upper limits on b and c as 10, and we want to maximize the value of a squared.&nbsp;This can be thought of as a right-angled triangle, and the value of a will be maximum when both b and c are equal to 10, giving&nbsp;$a^2=200$, but this would not give an integer value of aWe need to adjust&nbsp;$a^2$ to the biggest square less than 200, which is 196Giving the value of $a$ as 14.&nbsp;Therefore, 14 is the correct answer.&nbsp;&nbsp;

Squaring on both sides, we get:$x+6\sqrt{\ 2}+x-6\sqrt{\ 2}-2\left(x^2-72\right)^{\frac{1}{2}}=8$$x-\left(x^2-72\right)^{\frac{1}{2}}=4$ Bringing x to the other side, we get:$-\left(x^2-72\right)^{\frac{1}{2}}=4-x$Squaring on both sides again, we get:$x^2-72=16+x^2-8x$$8x=88$$x=11$Therefore, 11 is the correct answer.&nbsp;

Opening the square on the left-hand side, we get&nbsp;$a^2+3b^2+2ab\sqrt{\ 3}$Comparing the rational part on both side,s we get:&nbsp;$a^2+3b^2=52$And comparing the irrational par,t we get:&nbsp;$2ab\sqrt{\ 3}=30\sqrt{\ 3}$$ab=15$, Since we are given that a and b are natural numbers, the possible values of a and b are (1,15), (3, 5), (5, 3), or (15,1)Putting these values in the first relation we got, we see that 15 squared would exceed the required value and would not be the case.&nbsp;We need not check if a=5, b=3 or a=3, b=5 since the answer would be the same.&nbsp;(a=5 and b=3 would satisfy it)a+b = 5+3 = 8Therefore, Option B is the correct answer.&nbsp;

Taking&nbsp;$10^x=a$we get&nbsp;$a+\frac{4}{a}=\frac{81}{2}$This would give the quadratic equation:&nbsp;$2a^2-81a+8=0$We want to find the sum of possible values of x, let the value of x be x1 and x2these would correspond to log a1, and log a2The sum of log a1 + log a2 would be log (a1 x a2)From the quadratic equation we got above, we can see that the product of the possible values of a would-be 8/2 = 4Threfore, the sum of values of x would be log (4) which would be&nbsp;$2\ \log_{10}2$Therefore, Option&nbsp;A is the correct answer.&nbsp;

Taking a log for each of the expressions, we get the following:$\log_34=a,\ \log_45=b,\ \log_56=c,\ \log_67=d,\ \log_78=e,\ \log_89=f$The expression $abcef$ would then be:&nbsp;$\log_34\times\ \log_45\times\ \log_56\times\ \log_67\times\ \log_78\times\ \log_89$Next, we can use this property of log:&nbsp;$\frac{\log_ba}{\log_bc}=\log_ca$Using this, we get:$\frac{\log\ 4}{\log\ 3}\times\ \frac{\log\ 5}{\log\ 4}\times\ \frac{\log\ 6}{\log\ 5}\times\ \frac{\log\ 7}{\log\ 6}\times\ \frac{\log\ 8}{\log\ 7}\times\ \frac{\log\ 9}{\log\ 8}$All the terms will cancel out except:&nbsp;$\frac{\log\ 9}{\log\ 3}=\log_39=2$Therefore, 2 is the correct answer.&nbsp;

Given:$ \log_{x}(x^2 + 12) = 4 $$ x^2 + 12 = x^4 $$ x^4 - x^2 - 12 = 0 $$ x^4 - 4x^2 + 3x^2 - 12 = 0 $$ x^2(x^2 - 4) + 3(x^2 - 4) = 0 $$ (x^2 - 4)(x^2 + 3) = 0 \implies$ since $x$ is a positive real number, $x = 2$Now, given $ 3 \log_{y} x = 1 $$ \log_{y} x = \frac{1}{3} $$ x = y^{\frac{1}{3}} $$ y = x^3 $$ y = 8 $$ x + y = 2 + 8 = 10 $

Given,$ \sqrt{5x+9} + \sqrt{5x - 9} = 3(2 + \sqrt{2}) $$ \sqrt{5x+9} + \sqrt{5x-9} = 6 + 3\sqrt{2} $$ \sqrt{5x+9} + \sqrt{5x-9} = \sqrt{36} + \sqrt{18} $Comparing the L.H.S. and R.H.S.$ 5x+9=36 $ =&gt; $ 5x=27 $ =&gt; $ x=\dfrac{27}{5} $ (can be verified using the second term as well).$ \sqrt{10x+9} = \sqrt{\left(10\times\dfrac{27}{5}\right)+9} = \sqrt{63}=3\sqrt{7} $

It is given that $ \log_{\sqrt{3}}(x) + \frac{\log_3(25)}{\log_3(0.008)} = \frac{16}{3} $, which can be written as:=&gt; $ 2 \log_3 x + \log_{0.008} 25 = \frac{16}{3} $=&gt; $ 2 \log_3 x + \log_{\frac{8}{1000}} 25 = \frac{16}{3} $=&gt; $ 2 \log_3 x + \log_{\frac{1}{125}} 25 = \frac{16}{3} $=&gt; $ 2 \log_3 x + \log_{5^{-3}} (5)^2 = \frac{16}{3} $=&gt; $ 2 \log_3 x - \frac{2}{3} = \frac{16}{3} $=&gt; $ 2 \log_3 x = \frac{16}{3} + \frac{2}{3} $=&gt; $ 2 \log_3 x = 6 $=&gt; $ \log_3 x^2 = 6 \Rightarrow x^2 = 3^6 $Hence, $ \log_3 \left(3 \cdot x^2\right) = \log_3 \left(3 \cdot 3^6\right) = \log_3 3^7 = 7 $

It is given that $ x^8 + \left(\frac{1}{x}\right)^8 = 47 $, which can be written as:=&gt; $ \left(x^4\right)^2 + \left(\frac{1}{x^4}\right)^2 = 47 $=&gt; $ \left(x^4 + \frac{1}{x^4}\right)^2 - 2 \cdot x^4 \cdot \frac{1}{x^4} = 47 $=&gt; $ \left(x^4 + \frac{1}{x^4}\right)^2 = 49 $=&gt; $ x^4 + \frac{1}{x^4} = 7 $Similarly, $ x^4 + \frac{1}{x^4} = 7 $ can be expressed as:=&gt; $ \left(x^2\right)^2 + \left(\frac{1}{x^2}\right)^2 = 7 $=&gt; $ \left(x^2 + \frac{1}{x^2}\right)^2 - 2 \cdot x^2 \cdot \frac{1}{x^2} = 7 $=&gt; $ \left(x^2 + \frac{1}{x^2}\right)^2 = 9 $=&gt; $ x^2 + \frac{1}{x^2} = 3 $By the same logic, we get $ x + \frac{1}{x} = \sqrt{5} $Now, $ x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3 \cdot x \cdot \frac{1}{x}\left(x + \frac{1}{x}\right) $=&gt; $ x^3 + \frac{1}{x^3} = \left(\sqrt{5}\right)^3 - 3\sqrt{5} = 2\sqrt{5} $By the same logic, we can say that=&gt; $ x^9 + \frac{1}{x^9} = \left(x^3 + \frac{1}{x^3}\right)^3 - 3 \cdot x^3 \cdot \frac{1}{x^3}\left(x^3 + \frac{1}{x^3}\right) $=&gt; $ x^9 + \frac{1}{x^9} = \left(2\sqrt{5}\right)^3 - 3\left(2\sqrt{5}\right) $=&gt; $ x^9 + \frac{1}{x^9} = 40\sqrt{5} - 6\sqrt{5} = 34\sqrt{5} $

$\left(\sqrt{\frac{7}{5}}\right)^{3x - y} = \frac{875}{2401}$$\left(\frac{7}{5}\right)^{\frac{3x - y}{2}} = \frac{125}{343}$$\left(\frac{7}{5}\right)^{\frac{3x - y}{2}} = \left(\frac{7}{5}\right)^{-3}$3x − y = −6$\left(\frac{4a}{b}\right)^{6x - y} = \left(\frac{2a}{b}\right)^{y - 6x}$Therefore, y = 6x as the bases are different so the power should be zero for the results to be equal.3x − y = −6or, 3x − 6x = −6or x = 2y = 6x = 12

$5 - \log_{10}\sqrt{1+x} + 4\log_{10}\sqrt{1-x} = \log_{10}\frac{1}{\sqrt{1-x^2}}$We can re-write the equation as: $5 - \log_{10}\sqrt{1+x} + 4\log_{10}\sqrt{1-x} = \log_{10}\left(\sqrt{1+x}\times\sqrt{1-x}\right)^{-1}$$5 - \log_{10}\sqrt{1+x} + 4\log_{10}\sqrt{1-x} = (-1)\log_{10}(\sqrt{1+x}) + (-1)\log_{10}(\sqrt{1-x})$$5 = -\log_{10}\sqrt{1+x} + \log_{10}\sqrt{1+x} - \log_{10}\sqrt{1-x} - 4\log_{10}\sqrt{1-x}$$5 = -5\log_{10}\sqrt{1-x}$$\sqrt{1-x} = \frac{1}{10}$Squaring both sides: $(\sqrt{1-x})^2 = \frac{1}{100}$$\therefore\ x = 1 - \frac{1}{100} = \frac{99}{100}$Hence, $100x = 100 \times \frac{99}{100}$ = 99

We have$\log_2 \{ 3 + \log_3 \{ 4 + \log_4 (x - 1) \} \} = 2$we get&nbsp;&nbsp;$3 + \log_3 \{ 4 + \log_4 (x - 1) \} = 4$we get&nbsp;&nbsp;$\log_3 (4 + \log_4 (x - 1)) = 1$we get&nbsp;&nbsp;$4 + \log_4 (x - 1) = 3$$\log_4 (x - 1) = -1$$x - 1 = 4^{-1}$$x = \frac{1}{4} + 1 = \frac{5}{4}$4x = 5

$\frac { \log _ { 15 } ^ { a } + \log _ { 32 } ^ { a } } { \left( \log _ { 15 } ^ { a } \times \log _ { 32 } ^ { a } \right) }$ = 4we get&nbsp;$\log⁡_a$(log⁡ 32&nbsp;+ log⁡&nbsp;15) = 4(log⁡&nbsp;a$)^2$we get&nbsp;(log⁡ 32&nbsp;+log⁡&nbsp;15)=4log⁡a= log⁡480 = $log⁡ a^4$= $a^4$&nbsp;= $480$so we can say a is between 4 and 5 .

$2^{Y^2 (\log_3 5)} = 5^{Y^2 (\log_3 2)}$Given,&nbsp;&nbsp;$5^{Y^2 (\log_3 2)} = 5^{(\log_2 3)}$⇒ $Y^2 (\log_3 2) = (\log_2 3)$&nbsp;&nbsp;⇒ $Y^2 = (\log_2 3)^2$⇒ $Y = (-\log_2 3)$ or $(\log_2 3)$Since $Y$ is a negative number,&nbsp;&nbsp;$Y = (-\log_2 3) = (\log_2 \tfrac{1}{3})$

$\log_a 30 = A \ \text{ or }\ \log_a 5 + \log_a 2 + \log_a 3 = A \quad ...(1)$$\log_a \left(\frac{5}{3}\right) = -B \ \text{ or }\ \log_a 3 - \log_a 5 = B \quad ...(2)$$\text{and finally } \log_a 2 = 3$$\text{Substituting this in (1) we get } \log_a 5 + \log_a 3 = A - 3$$\text{Now we have two equations in two variables (1) and (2). On solving we get}$$\log_a 3 = \frac{A + B - 3}{2} \quad \text{ or }\ \log_3 a = \frac{2}{A + B - 3}$

CAT Logarithms

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

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Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

Question 13.

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Question 15.

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Free CAT Quant Questions

CAT Logarithms

Question 1.

<div><p>If x is a positive real number such that $4 \log_{10} x + 4 \log_{100} x + 8 \log_{1000} x = 13$, then the greatest integer not exceeding x, is</p></div>

Question 2.

<div><p>If $(a + b \sqrt{n})$ is the positive square root of $(29 - 12\sqrt{5})$, where a and b are integers, and n is a natural number, then the maximum possible value of $(a + b + n)$ is</p></div>

Question 3.

<div><p>If a, b and c are positive real numbers such that $a &gt; 10$ $\geq b$$ \geq c$ and $\cfrac{\log_8 (a + b)}{\log_2c} + \cfrac{\log_{27} (a - b)}{\log_3c} = \cfrac{2}{3}$, then the greatest possible integer value of a is</p></div>

Question 4.

<div><p>If $(x + 6\sqrt{2})^{\cfrac{1}{2}} - (x - 6\sqrt{2})^{\cfrac{1}{2}} = 2\sqrt{2}$, then x equals</p></div>

Question 5.

<div><p>If $(a + b\sqrt{3})^2 = 52 + 30\sqrt{3}$, where a and b are natural numbers, then $a + b$ equals</p></div>

Question 6.

<div><p>The sum of all distinct real values of x that satisfy the equation $10^x + \cfrac{4}{10^x} = \cfrac{81}{2}$, is</p></div>

Question 7.

<div><p>If $3^a = 4, 4^b = 5, 5^c = 6, 6^d = 7, 7^e = 8$ and $8^f = 9$, then the value of the product abcdef is</p></div>

Question 8.

<div><p>If x and y are positive real numbers such that $\log _x\left(x^2+12\right)=4$ and $3 \log _y x=1$ , then x+y equals</p></div>

Question 9.

<div><p>If $\sqrt{5 x+9}+\sqrt{5 x-9}=3(2+\sqrt{2})$, then $\sqrt{10 x+9}$ is equal to</p></div>

Question 10.

<div><p>For some positive real number x , if $\log _{\sqrt{3}}(x)+\frac{\log _x(25)}{\log _x(0.008)}=\frac{16}{3}$, then the value of $\log _3\left(3 x^2\right)$ is</p></div>

Question 11.

<div><p>If x is a positive real number such that $x^8+\left(\frac{1}{x}\right)^8=47$, then the value of $x^9+\left(\frac{1}{x}\right)^9$ is</p></div>

Question 12.

<div><p>If $\left(\sqrt{\frac{7}{5}}\right)^{3 x-y}=\frac{875}{2401}$ and $\left(\frac{4 a}{b}\right)^{6 x-y}=\left(\frac{2 a}{b}\right)^{y-6 x}$, for all non-zero real values of a and b, then the value of x + y is</p></div>

Question 13.

<div><p>If 5 - $\log _{10} \sqrt{1+x}$ + 4$\log _{10} \sqrt{1-x}$ = $\log _{10} \frac{1}{\sqrt{1-x^{2}}}$, then 100 x equals</p></div>

Question 14.

<div><p>If $log_{2}$[3 + $log_{3}${4 + $log_{4}$(x - 1)}] - 2 = 0 then 4x equals</p></div>

Question 15.

<div><p>For a real number a, if $\frac{\log _{15} a+\log _{32} a}{\left(\log _{15} a\right)\left(\log _{32} a\right)}=$ = 4 then a must lie in the range</p></div>

Question 16.

<div><p>If y is a negative number such that $2^{y^{2}log_{3}5}$ = $5^{log_{2}3}$, then y equals</p></div>

Question 17.

<div><p>Let $log_{a}$30 = A, $log_{a}\frac{5}{3}$ = -B and $log_{2}$a = $\frac{1}{3}$ , then $log_{3}$a equals</p></div>

Question 18.

<div><p>The number of distinct integers $n$ for which $\log_{\left(\frac{1}{4}\right)}(n^2 - 7n + 11) $&gt; $0$, is</p></div>

Question 19.

<div><p>If $\log_{64}x^{2}+\log_{8}\sqrt{y}+3\log_{512}(\sqrt{y}z)=4$ where x, y and z are positive real numbers, then the minimum possible value of $(x+y+z)$ is</p></div>

Question 20.

<div><p>The sum of all possible real values of x for which $\log_{x-3}(x^{2}-9)=\log_{x-3}(x+1)+2,$ is</p></div>

Free CAT Quant Questions