SSC > Logarithms

Explanation

We have log3 log4 x > 0

=> {tex} log4x > 1=> x > 4 {/tex}

=> x > 4

Explanation

{tex} log_{\frac{1}{4}} \left(a^{2}-1\right) < log_{ \left(\frac{1}{4}\right) } \left(a+1\right)^{2} {/tex}

=> {tex} a^{2}-1 > \left(a+1\right)^{2} {/tex}

=> {tex} a^{2}-1 > a^{2}+1+2a {/tex}

=> {tex} 2a + 2 < 0 => a < -1 {/tex}

Explanation

{tex} log \left(\frac{a+b}{2}\right) = \frac{1}{2} \left(log_{e}a + log_{e}b\right) {/tex}

=> {tex} \frac{a+b}{2} = \sqrt{ab} {/tex}

=> {tex} a+b-2\sqrt{ab} = 0 {/tex}

=> {tex} \sqrt{a} = \sqrt{b} => a = b {/tex}

Explanation

By definition x ≠ 1, =1

Given equation can be written as

{tex} log \frac{ \left(x+1\right)^{2} }{x^{2}-1} = log2 => \frac{x+1}{x-1}=2 => x = 3{/tex}

Explanation

Let {tex} x = log 2^{7} {/tex}

=> {tex} 2^{x} = 7 {/tex}

which is only possible for irrational number

Explanation

{tex} y = 2^{\frac{1}{log}x \left(8\right) } {/tex}

=> {tex} y = 2 ^{log 8x} => y = 2^{log 2\sqrt[3]{x}} {/tex}

=> {tex} x = y^{3} {/tex}

Explanation

{tex} log\ 0.5 \sin x + log\ 0.5 \cos x = 1 {/tex}

=> {tex} log\ 0.5 \sin x \cos x = 1 {/tex}

=> {tex} \sin x \cos x = \frac{1}{2} => \sin 2x = 1 {/tex}

=> {tex} 2x = n \pi + \left(-1\right)^{n}\frac{ \pi }{2} => x = \frac{n \pi }{2}+ \left(-1\right)^{n}\frac{ \pi }{4} {/tex}

Since {tex} log_{0.5} \sin x\ and\ log_{0.5}.\cos x\ are\ real {/tex}

{tex} \therefore \ \sin x \ and \ \cos x \ must \ lie\ in\ first\ quardrant {/tex}

{tex} \therefore x = \frac{ \pi }{4}, -2 \pi + \frac{ \pi }{4} {/tex}

Hence, the number of solutions is 2.

Explanation

{tex} log_{ax}x = \frac{1}{log_{x}ax} => log_{ax}x = \frac{1}{1+log_{x}a}{/tex}

If a,b,c are in GP

=> {tex} log_{x}a, log_{x}b, log_{x}c\ are\ in\ AP {/tex}

=> {tex} 1+log_{x}a,1+log_{x}b,1+log_{x}c\ are\ in\ AP {/tex}

=> {tex} \frac{1}{1+log_{x}a}, \frac{1}{1+log_{x}b}, \frac{1}{1+log_{x}c}\ are\ in\ HP {/tex}

=> {tex} log_{ax}x,\ log_{bx}x\ log_{cx}x\ are\ in\ HP. {/tex}

Explanation

{tex} alog_{5} \left(x^{2}-11x+43\right)<2 {/tex} and {tex} x^{2}-11x+43>0 {/tex}
=> {tex} x^{2}-11x+43 < 5^{2}{/tex} and {tex} \left(x - \frac{11}{2}\right)^{2}+\frac{51}{4}>0 {/tex}
=> {tex} x^{2}-11x+18<0 {/tex} and {tex} \left(x-2\right) \left(x-9\right)<0 {/tex}
Solution is {tex} \left(2, 9\right) {/tex}
For domain {tex} |x, -1| ≠ 0, ≠ -1 {/tex}
Now {tex} |x-1| = 1, => x - 1 = \pm 1 {/tex}
=> x = 0, 2
x = 0 is not in the domain and x = 2 satisfies the given equation.
If x-1=> 0 i.e., x => 1 then the given equation becomes
{tex} 2 log_{3}x - \frac{4}{log_{3}x}=7 => x = 81, \frac{1}{\sqrt{3}} {/tex}
But {tex} \frac{1}{\sqrt{3}} {/tex} being less than 1 is not valid
Hence, x = 2, 81

Explanation

{tex} log \cos x \sin x \ge 2 => \sin x \le \cos^{2}x {/tex}

=>{tex} sinx \le 1 - sin^{2}x {/tex}

=> {tex} \sin^{2}x + \sin x - 1 \le 0 {/tex}

=> {tex} \left(\sin x + \frac{1}{2}\right)^{2} - \frac{5}{4} \le 0 {/tex}

Also by definition of logarithm

{tex} \sin x > 0, \cos x > 0, \cos x ≠ 1 {/tex}

=> {tex} \sin x + \frac{1}{2} \le \frac{\sqrt{5}}{2} {/tex}

=> {tex} 0 < \sin x \le \frac{\sqrt{5}-1}{2} {/tex}

Explanation

{tex} log^{2}xy \ge 6 => xy \ge 2^{6} {/tex}

Now {tex} \frac{x + y}{2} \ge \sqrt{xy} {/tex}

{tex} Now \frac{x+y}{2} \ge 2^{3} => x+y \ge 2^{4} => x+y \ge 16 {/tex}

Explanation

{tex} 4^{log_{3}\frac{31}{2}} + 9^{log_{2}2^{2}} = 10^{log_{x}83} {/tex}

=> {tex} 4^{\frac{1}{2}} + 9^{2} = 10^{log}x^{83} {/tex}

=> {tex} 2 + 81 = 10^{log}x^{83} {/tex}

=> {tex} 83 = 10^{log}x^{83} {/tex}

=> x = 10

Explanation

(A) {tex} \frac{1}{log_{3} \pi } + \frac{1}{log_{4} \pi } = log_{ \pi }3 + log_{ \pi }4 = log_{ \pi} 12 > 2{/tex}

{tex} 12 > \pi ^{2} {/tex}

(B) {tex} log_{3}5 {/tex} is an irrational number

(C) {tex} log\sqrt{8}x = \frac{10}{3} => x = \left(\sqrt{8}\right)^{\frac{10}{3}} = 2^{5} {/tex}

{tex} x = 32 {/tex}

(D) {tex} log_{x} \left(a^{2}+1\right) < 0, a ≠ 0 => a^{2}+1 > 1 {/tex}

So, log is negative

Hence, base is (0, 1).

Explanation

From the given relation, we have

{tex} a = y^{1-log ax} = z^{1-log ay} {/tex}

{tex} \therefore\ log_{a}a = \left(1-log_{a}x\right)log_{a}y {/tex}

{tex} and\ log_{a}a = \left(1-log_{a}y\right)log_{a}z {/tex}

=> {tex} log_{a}y \left(1-log_{a}x\right) = 1 {/tex}

{tex} and\ log_{a}z \left(1-log_{a}y\right) = 1 {/tex}

=> {tex} log_{a}y = \frac{1}{1-log_{a}x} {/tex}

{tex} and\ log_{a}z = \frac{1}{1-log_{a}y} {/tex}

{tex} \therefore log_{a}z = \frac{1}{1-log_{a}y} = \frac{1}{1-\frac{1}{1-log_{a}x}}{/tex}

= {tex} \frac{1-log_{a}x}{-log_{a}x} {/tex}

{tex} Now \frac{1}{1-log_{a}z} = \frac{1}{1 + \frac{1-log_{a}x}{logt_{a}x}}= log_{a}x {/tex}

{tex} log_{a}x = \frac{1}{1 - log_{a}z} {/tex}

{tex} x = a \frac{1}{1-log_{a}z} {/tex}

Explanation

Series {tex} 1+\frac{3}{4}+ \left(\frac{3}{4}\right)^{2}+...+nterms=\frac{1- \left(\frac{3}{4}\right)^{n} }{1-\frac{3}{4}} {/tex}

and {tex} 1+\frac{3}{4}+ \left(\frac{3}{4}\right)^{2}+...=\frac{1}{1-\frac{3}{4}}=4 {/tex}

Now, {tex} \frac{1- \left(\frac{3}{4}\right)^{n} }{\frac{1}{4}}=4-10^{-6} {/tex}

=> {tex} 1- \left(\frac{3}{4}\right)^{n}=1-\frac{1}{4} \left(10^{-6}\right) {/tex}

=> {tex} \left(\frac{3}{4}\right)^{n}=\frac{10^{-6}}{4} {/tex}

=> {tex} nlog_{10}\frac{3}{4}=log_{10}10^{-6}-log_{10}4 {/tex}

=> {tex} n \left(0.47712-2 \times 0.30103\right)=-6-2 \times \left(0.30103\right) {/tex}

=> {tex} n=\frac{6.6026}{0.12494}=53 {/tex}

Explanation

Given that, {tex} x^{log}x^{2}=2=log_{3} \left(x+y\right) {/tex}

=> {tex} x+y=9 {/tex} and {tex} x^{2}+y^{2}=65 {/tex}

=> {tex} x=8, y=1 or x=1, y=8 {/tex}

But x≠ 1

Therefore, x = 8, y = 1

Explanation

{tex} log_{a}n \ log_{b}n+log_{b}n \ log_{c}n+log_{c}n \ log_{a}n {/tex}

={tex} \frac{1}{log_{n}a \ log_{n}b}+\frac{1}{log_{n}b \ log_{n}c}+\frac{1}{log_{n}c \ log_{n}a} {/tex}

{tex} \left[ \therefore log_{m}n=\frac{1}{log_{n}m} \right] {/tex}

= {tex} \frac{log_{n}c+log_{n}a+log_{n}b}{log_{n}a \ log_{n}b \ log_{n}c} {/tex}

= {tex} \frac{log_{n} \left(abc\right) }{log_{n}a \ log_{n}b \ log_{n}c}=\frac{log_{a}n \ log_{b}n \ log_{c}n}{log_{abc}n} {/tex}

Explanation

Here, {tex} 2log_{10}x-log_{x} \left(10\right)^{-2}=2log_{10}x+2log_{x}10 {/tex}

= {tex} 2log_{10}x+2\frac{1}{log_{10}x} {/tex}

= {tex} 2\{ log_{10}x+\frac{1}{log_{10}x} \} {/tex}

Using {tex} AM \ge GM {/tex} we get

{tex} \frac{log_{10}x+\frac{1}{log_{10}x}}{2}\ge \left(log_{10}x\frac{1}{log_{10}x}\right)^{\frac{1}{2}} {/tex}

=> {tex} log_{10}x+\frac{1}{log_{10}x}\ge 2 {/tex}

{tex} \therefore 2log_{10}x-log_{x} \left(0.01\right)\ge 4 {/tex}

Therefore, Least value is 4.

Explanation

{tex} log_{10}1000 {/tex} =

= {tex} log_{10}10^{3} {/tex}

={tex} 3log_{10}10 = 3 \times 1 = 3 {/tex}

Explanation

Simplify: {tex} log_{10}2500 {/tex}

{tex} log_{10}2500 {/tex} = {tex} log_{10} \left(25 \times 100\right) {/tex}

={tex} log_{10}25+log_{10}100 {/tex}

={tex} log_{10}5^{2}+log_{10}10^{2} {/tex}

={tex} 2log_{10}5+2log_{10}10 {/tex}

={tex} 2log_{10}5+2 {/tex}

Therefore, {tex} \left[ log_{10}10=1 \right] {/tex}

Explanation

{tex} log_{10}2500 - log_{10}25 {/tex}

={tex} log_{10} \left(\frac{2500}{25}\right) {/tex}

={tex} log_{10}100 {/tex}

={tex} log_{10}10^{2} = 2log_{10}10 = 2 {/tex}

Explanation

{tex} log_{27}81=\frac{4}{3} {/tex}

{tex} 81= \left(27\right)^{\frac{4}{3}}= \left(3^{3}\right)^{\frac{4}{3}} {/tex}

= {tex} 3^{4} {/tex}

Explanation

{tex} log_{10}4+log_{10}25 {/tex}

{tex} log_{10} \left(4 \times 25\right)=log_{10}1000 {/tex}

={tex} log_{10}10^{2} {/tex}

={tex} 2log_{10}10 {/tex}

=2

Explanation

{tex} log_{3}72-log_{3}8 = log_{3} \left(9 \times 8\right)-log_{3}8 {/tex}

={tex} log_{3}9+log_{3}8-log_{3}8 {/tex}

={tex} log_{3}9 {/tex}

={tex} log_{3}3^{2}=2log_{3}3 {/tex}

=2

Explanation

{tex} 3log_{a}2-log_{a}8=log_{a}2^{3}-log_{a}8 {/tex}

={tex} log_{a}2^{3}-log_{a}8 {/tex}

={tex} log_{a}8-log_{a}8=0 {/tex}