Question

solve for x : (i) (1/7)^4-2x=root 7
please answer is x=9/4 but I need full solution
please guys be fast​

Answer 1

ANSWER:

Given:

$\:\:\bullet\:\:\left(\dfrac{1}{7}\right)^{4-2x}=\sqrt7$

To Find:

• Value of x

Solution:

We are given that,

$\implies\left(\dfrac{1}{7}\right)^{4-2x}=\sqrt7$

Taking (-) common from the power in LHS,

$\implies\left(\dfrac{1}{7}\right)^{4-2x}=\sqrt7$

$\implies\left(\dfrac{1}{7}\right)^{-(-4+2x)}=\sqrt7$

$\implies\left(\dfrac{1}{7}\right)^{-(2x-4)}=\sqrt7$

We know that,

$\hookrightarrow a^{-x}=\left(\dfrac{1}{a}\right)^x$

So,

$\implies\left(\dfrac{1}{7}\right)^{-(2x-4)}=\sqrt7$

$\implies 7^{2x-4}=\sqrt7$

We know that,

$\hookrightarrow \sqrt{x}=x^{\frac{1}{2}}$

So,

$\implies 7^{2x-4}=\sqrt7$

$\implies 7^{2x-4}=7^{\frac{1}{2}}$

As, the bases are same, we compare the powers,

$\implies 7^{2x-4}=7^{\frac{1}{2}}$

$\implies 2x-4=\dfrac{1}{2}$

Transposing 2 from RHS to LHS,

$\implies 2(2x-4)=1$

So,

$\implies 4x-8=1$

Transposing 8 to RHS,

$\implies 4x=1+8$

$\implies 4x=9$

Transposing 4 to RHS,

$\implies\bf x=\dfrac{9}{4}$

Therefore, value of x is 9/4.