Question

Find the value of x:(3/5)^2*(5/3)^2x=(125/27)
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Answer 1

Given:

The equation:-

$\bf \bigg( \dfrac{3}{5} \bigg)^2 \times \bigg( \dfrac{3}{5} \bigg)^{2x} = \dfrac{125}{27}$

What To Find:

We have to find the value of x.

How To Find:

To find the value of x,

• Make the same bases on both sides.
• Form an equation for the exponents.

Solution:

• Making the bases the same.

$\sf \Rightarrow \bigg( \dfrac{3}{5} \bigg)^2 \times \bigg( \dfrac{3}{5} \bigg)^{2x} = \dfrac{125}{27}$

125 can be written as 5³,

$\sf \Rightarrow \bigg( \dfrac{3}{5} \bigg)^2 \times \bigg( \dfrac{3}{5} \bigg)^{2x} = \dfrac{5^3}{27}$

27 can be written as 3³,

$\sf \Rightarrow \bigg( \dfrac{3}{5} \bigg)^2 \times \bigg( \dfrac{3}{5} \bigg)^{2x} = \dfrac{5^3}{3^3}$

Also written as,

$\sf \Rightarrow \bigg( \dfrac{3}{5} \bigg)^2 \times \bigg( \dfrac{3}{5} \bigg)^{2x} = \bigg( \dfrac{5}{3} \bigg) ^3$

Using $\sf a^{-n} = \dfrac{1}{a^{n}}$,

$\sf \Rightarrow \bigg( \dfrac{3}{5} \bigg)^2 \times \bigg( \dfrac{3}{5} \bigg)^{2x} = \bigg( \dfrac{3}{5} \bigg) ^{-3}$

Now the bases are the same.

• Solving the exponents.

⇒ 2 + 2x = -3

Take 2 to RHS,

⇒ 2x = -3 - 2

Subtract them,

⇒ 2x = -5

Take 2 to RHS,

$\sf \Rightarrow x = \dfrac{-5}{2}$

Divide them,

⇒ x = -2.5

∴ Hence, the value of x is -2.5.

Answer 2

Given:

The equation:-

$\bf \bigg( \dfrac{3}{5} \bigg)^2 \times \bigg( \dfrac{3}{5} \bigg)^{2x} = \dfrac{125}{27}$

What To Find:

We have to find the value of x.

How To Find:

To find the value of x,

Make the same bases on both sides.

Form an equation for the exponents.

Solution:

●Making the bases the same.

$\sf \Rightarrow \bigg( \dfrac{3}{5} \bigg)^2 \times \bigg( \dfrac{3}{5} \bigg)^{2x} = \dfrac{125}{27}$

125 can be written as 5³,

$\sf \Rightarrow \bigg( \dfrac{3}{5} \bigg)^2 \times \bigg( \dfrac{3}{5} \bigg)^{2x} = \dfrac{5^3}{27}$

27 can be written as 3³,

$\sf \Rightarrow \bigg( \dfrac{3}{5} \bigg)^2 \times \bigg( \dfrac{3}{5} \bigg)^{2x} = \dfrac{5^3}{3^3}$

Also written as,

$\sf \Rightarrow \bigg( \dfrac{3}{5} \bigg)^2 \times \bigg( \dfrac{3}{5} \bigg)^{2x} = \bigg( \dfrac{5}{3} \bigg) ^3$

Using $\sf a^{-n} = \dfrac{1}{a^{n}}$,

$\sf \Rightarrow \bigg( \dfrac{3}{5} \bigg)^2 \times \bigg( \dfrac{3}{5} \bigg)^{2x} = \bigg( \dfrac{3}{5} \bigg) ^{-3}$

Now the bases are the same.

Solving the exponents.

⇒ 2 + 2x = -3

Take 2 to RHS,

⇒ 2x = -3 - 2

Subtract them,

⇒ 2x = -5

Take 2 to RHS,

$\sf \Rightarrow x = \dfrac{-5}{2}$

Divide them,

⇒ x = -2.5

∴ Hence, the value of x is -2.5.