Question

$( \sqrt{ \frac{3}{5} } ) ^{x + 1} = \frac{125}{27}$
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Answer 1

Answer:

$\large\bold\red{x=-7}$

Step-by-step explanation:

Given,

${( \sqrt{ \frac{3}{5} } )}^{(x + 1)} = \frac{125}{27}$

Further Simplifying,

We get,

$= > {( \frac{3}{5} )}^{ \frac{x + 1}{2} } = \frac{ {5}^{3} }{ {3}^{3} } \\ \\ = > {( \frac{3}{5} )}^{ \frac{x + 1}{2} } = {( \frac{5}{3} )}^{3} \\ \\ = > {( \frac{3}{5}) }^{ \frac{x + 1}{2} } = \frac{1}{ ({ \frac{3}{5} )}^{3} } \\ \\ = > {( \frac{3}{5} )}^{ \frac{x + 1}{2} } \times {( \frac{3}{5}) }^{3} = 1 \\ \\ = > {( \frac{3}{5}) }^{3 + \frac{x + 1}{2} } = {( \frac{3}{5}) }^{0}$

Now,

The bases are same on both sides,

Therefore,

The exponent will also be same,

Therefore,

We get,

$= > 3 + \frac{x + 1}{2} = 0 \\ \\ = > 6 + x + 1 = 0 \\ \\ = > x + 7 = 0 \\ \\ = > x = - 7$

Hence,

x = -7 is the solution .