Question

Exponential form of 3(log7-log5)

Answer 1

3(log7 - log5) 
= 3(log7/5) [ ∵ logM - logN = log M/N]
= log(7/5)³  [ ∵ Mlog n = log n^M]
Assume that
log(7/5)³ to the base `e` =  x
⇒e^x = (7/5)³

Answer 2

Answer:

The required exponential form is $e^x=(\frac{7}{5})^3$

Step-by-step explanation:

Given : Expression $3(\log 7-\log 5)$

To find : Write the expression in exponential form ?

Solution :

Let $x=3(\log 7-\log 5)$

Applying logarithmic divisible rule,

$\log (\frac{A}{B})=\log A-\log B$

Here, A=7 and B=5

$x=3(\log (\frac{7}{5}))$

We know, $x\log a=\log a^x$

$x=\log (\frac{7}{5})^3$

Taking exponential both side,

$e^x=(\frac{7}{5})^3$

Therefore, The required exponential form is $e^x=(\frac{7}{5})^3$