Question

Find the value of k: (−4/5 )^2​ ​ x (4/5)^5​ ​ = (4/5)^(6​k+1)

Answer 1

Answer:

k=1

Step-by-step explanation:

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Answer 2

The value of k = 1

Given :

$\displaystyle \sf{ {\bigg( - \frac{4}{5} \bigg)}^{2} \times {\bigg( \frac{4}{5} \bigg)}^{5} ={\bigg( \frac{4}{5} \bigg)}^{(6k + 1)} }$

To find :

The value of k

Solution :

Step 1 of 2 :

Write down the given equation

Here the given equation is

$\displaystyle \sf{ {\bigg( - \frac{4}{5} \bigg)}^{2} \times {\bigg( \frac{4}{5} \bigg)}^{5} ={\bigg( \frac{4}{5} \bigg)}^{(6k + 1)} }$

Step 2 of 2 :

Find the value of k

$\displaystyle \sf{ {\bigg( - \frac{4}{5} \bigg)}^{2} \times {\bigg( \frac{4}{5} \bigg)}^{5} ={\bigg( \frac{4}{5} \bigg)}^{(6k + 1)} }$

$\displaystyle \sf{ \implies {\bigg( \frac{4}{5} \bigg)}^{2} \times {\bigg( \frac{4}{5} \bigg)}^{5} ={\bigg( \frac{4}{5} \bigg)}^{(6k + 1)} }\: \: \: \bigg[ \: \because \: {( - a)}^{2} = {a}^{2} \bigg]$

$\displaystyle \sf{ \implies } {\bigg( \frac{4}{5} \bigg)}^{2 + 5} ={\bigg( \frac{4}{5} \bigg)}^{(6k + 1)} \: \: \bigg[ \because \: {a}^{m} \times {a}^{n} = {a}^{m + n} \bigg]$

$\displaystyle \sf{ \implies } {\bigg( \frac{4}{5} \bigg)}^{7} ={\bigg( \frac{4}{5} \bigg)}^{(6k + 1)}$

$\displaystyle \sf{ \implies } 7 = {(6k + 1)}$

$\displaystyle \sf{ \implies } 6k = 7 - 1$

$\displaystyle \sf{ \implies } 6k = 6$

$\displaystyle \sf{ \implies } k = 1$

Hence the required value of k = 1

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