Question

$( \frac{2}{3}) ^{ - x} + ( \frac{2}{3})^{ - 4} = ( \frac{2}{3})^{5}$
Find "x"



Please answer the following questions as soon as possible and I will mark you as Brainliest answer
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Answer 1

=> (2/3)^-x + (2/3)^-4 = (2/3)^5

=> (2/3)^-x + (3/2)^4 = (2/3)^5

=> (2/3)^-x = (32/243) - (81/16)

=> (2/3)^-x = -4.931

which is not possible,because any positive number has any power it's answer will also positive, but in this case we see negative value (-4.931) which is not correct...

ao that X doesn't exist..

hey bro.. this is correct ans..

Answer 2

Step-by-step explanation:

Given:

• $( \frac{2}{3}) ^{ - x} + ( \frac{2}{3 } ) ^{ - 4} = ( \frac{2}{3} ) ^{5}$

Solution:

We have

$( \frac{2}{3} ) ^{ - x} + ( \frac{2}{3} ) ^{ - 4} = ( \frac{2}{3} ) ^{ - 5} \\ = > ( \frac{3}{2} ) ^{x} + ( \frac{3}{2} ) ^{4} = ( \frac{3}{2} ) ^{5} \\ \\ = > ( \frac{3}{2} ) ^{x} + ( \frac{3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2} ) = ( \frac{3 \times 3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 2} ) \\ \\ = > ( \frac{3}{2}^{x} ) + ( \frac{81}{16} ) = ( \frac{243}{32} ) \\ \\ = > ( \frac{3}{2} ^{x} ) = \frac{243}{32} - \frac{81}{16} \: take \: lcm \\ \\ = > ( \frac{3}{2}) ^{x} = \frac{243 - 162}{32} \\ \\ = > ( \frac{3}{2} ^{x}) = \frac{81}{32} \\ \\ = > ( \frac{3}{2} ^{x} ) = ( \frac{3}{2} ) \: or \\ \\ = > ( \frac{2}{3} ^{ - x} ) = ( \frac{2}{3} ^{ - 1} ) \: bases \: are \: same \: so \: power \: wil \: be \: also \: equal \\ \\ = > ( - x) = ( - 1) \\ \\ = > x = 1$