Question

If log base x of (1/343)= -3 .then value of x=?

Answer 1

The value of x is "7".

Step-by-step explanation:

We have,

$\log_{x}\dfrac{1}{343} =-3$

To find, the value of x = ?

$\log_{x}\dfrac{1^{3} }{7^{3} } =-3$

⇒ $\log_{x}(\dfrac{1}{7} )^{3}  } =-3$

⇒$3 \log_{x}(\dfrac{1}{7})} =-3$

[ ∵ $\log a^{m}=m \log a$]

⇒$\log_{x}(\dfrac{1}{7})} =\dfrac{-3}{3} =-1$

⇒$\log_{x}(\dfrac{1}{7})} =-1$

⇒ $x^{-1} =\dfrac{1}{7}$

⇒ x = 7

Hence, the value of x is 7.

Answer 2

Given:

A logarithmic equation$log_{x}(\frac{1}{343} ) = -3$

To Find:

The value of x is?

Solution:

The given problem can be solved using the formulae of logarithms.

1. The formulae used in solving the given equation are,

• $log_{x}(a^n ) = nlog_{x}(a )$ ( Formula 1 )

• $log_{a}(a ) = 1$  ( Formula 2 )

• $1/a^{-n} = a^n$ ( Formula 3 )

2. The given equation is,

=>$log_{x}(\frac{1}{343} ) = -3$,

It can be also written as,

=>  $log_{x}(\frac{1}{7^3} ) = -3$

=>  $(log_{x}{7^{-3}} ) = -3$

=>  $-3(log_{x}{7^} ) = -3$

=>   $(log_{x}{7 ) = \frac{-3}{-3}$

=> $(log_{x}{7 ) = 1$

=> x = 7.

Therefore, the value of x is 7.