Question

x if (√4/5)^x- 1 =125/64​

Answer 1

Given :  (√4/5)ˣ⁻¹  = 125/65

$(\sqrt{\frac{4}{5}})^{x-1}=\dfrac{125}{64}$

To Find: Solve for x

Solution:

$(\sqrt{\frac{4}{5}})^{x-1}=\dfrac{125}{64}$

125 = 5³  = (√5)⁶

64 =  4³  = (√4)⁶

$\dfrac{125}{64} =\dfrac{(\sqrt{5})^6 }{\sqrt{4})^6 }$

$\dfrac{125}{64} =({\sqrt{\dfrac{5}{4}})^6$

$\dfrac{125}{64} =({\sqrt{\dfrac{4}{5}})^{-6}$

$(\sqrt{\dfrac{4}{5}})^{x-1}=\dfrac{125}{64}$

$(\sqrt{\dfrac{4}{5}})^{x-1}=({\sqrt{\dfrac{4}{5}})^{-6}$

Base are same so equate powers

Hence x - 1 =   - 6

=> x = - 5

Value of x is -5

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Answer 2

SOLUTION

GIVEN

$\displaystyle \sf{ { \bigg( \sqrt{ \frac{4}{5} } \bigg)}^{x - 1} = \frac{125}{64} }$

TO DETERMINE

The value of x

FORMULA TO BE IMPLEMENTED

We are aware of the formula on indices that :

‌$\sf{1. \: \: {a}^{m} \times {a}^{n} = {a}^{m + n} }$

‌$\displaystyle \sf{2. \: \: \: \frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n} }$

‌$\displaystyle \sf{3. \: \: \: { ({a}^{m} )}^{n} = {a}^{mn} }$

‌$\displaystyle \sf{4. \: \: {a}^{0} = 1}$

‌

EVALUATION

Here the given equation is

$\displaystyle \sf{ { \bigg( \sqrt{ \frac{4}{5} } \bigg)}^{x - 1} = \frac{125}{64} }$

We find the value of x as below

$\displaystyle \sf{ { \bigg( \sqrt{ \frac{4}{5} } \bigg)}^{x - 1} = \frac{125}{64} }$

$\displaystyle\sf{ \implies { \bigg[ { \bigg( \frac{4}{5}\bigg)}^{ \frac{1}{2} } \bigg]}^{x - 1} = \frac{ {5}^{3} }{ {4}^{3} } }$

$\displaystyle\sf{ \implies { \bigg[ { \bigg( \frac{4}{5}\bigg)}^{ \frac{1}{2} } \bigg]}^{x - 1} = { \bigg( \frac{5}{4}\bigg)}^{ 3 } }$

$\displaystyle\sf{ \implies { \bigg( \frac{4}{5}\bigg)}^{ \frac{x - 1}{2} } = { \bigg[ { \bigg( \frac{4}{5}\bigg)}^{ - 1} \bigg]}^{3} }$

$\displaystyle\sf{ \implies { \bigg( \frac{4}{5}\bigg)}^{ \frac{x - 1}{2} } = { \bigg( \frac{4}{5}\bigg)}^{ - 3} }$

$\displaystyle\sf{ \implies \frac{x - 1}{2} = - 3}$

$\displaystyle\sf{ \implies x - 1 = - 6}$

$\displaystyle\sf{ \implies x = - 5}$

FINAL ANSWER

Hence the required value of x = - 5

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