Question

show that 2 log 5/8 + log 128/125 + log 5/2 =0 ​

Answer 1

Given,

$\sf 2 log \dfrac{5}{8} + log \dfrac{128}{125} + log \dfrac{5}{2} =0$

Here,

$\longrightarrow\sf LHS=2\:log \:\dfrac{5}{8} + log \:\dfrac{128}{125} + log \:\dfrac{5}{2}$

$\longrightarrow\sf RHS=0$

Taking the LHS.

We know,

$\longrightarrow\sf log\:\dfrac{a}{b}=log\:a-log\:b$

So,

$=\sf 2\:log\:5-2\:log\:8+log\:128-log\:125+log\:5-log\:2$

$\sf=2\:log\:5-2\:log\:2^3+log\:2^7-log\:5^3+log\:5-log\:2$

$\sf= 3\:log\:5-4\:log\:2-7\:log\:2-3\:log\:5-3\:log\:2$

$=\sf 3\:log\:\dfrac{5}{5}+7\:log\:\dfrac{2}{2}$

$=\sf 3\:log\:1+7\:log\:1$

We know,

• log 1 = 0

$\sf=3\times0+7\times0$

$=\sf 0+0$

$=\sf 0$

$\sf=RHS$

Therefore,

$\sf LHS=RHS$

$\sf 2 log \dfrac{5}{8} + log \dfrac{128}{125} + log \dfrac{5}{2} =0$

Hence Proved !!!

Answer 2

Answer:

the answer is absolutely correct.