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Find the value of x in the following equation: 64^x+1 + 8^2x = 1040

Solving for x using the rules of indices

Answers

CHARLES
64^x+1 + 8^2x = 1040 can be written as;

8^2(x+1) + 8^2x = 1040

8^2x+2 + 8^2x = 1040

8^2x.8^2 + 8^2x = 1040 ……………………………….. (i)

Let 8^2x = y ………………………………………... (ii)

Then equation (i) becomes;

64y +y = 1040
65y = 1040
Therefore,
y = 1040/65 = 16

Substituting the value of y into equation (ii), we have;
8^2x = 16
2^3(2x) = 24
2^6x = 24 …………………………………………... (iii)

Equating the indices of equation (iii),
6x = 4
Therefore, the value of x is:
x = 4/6
= 2/3
Charcher answered the question on March 16, 2018 at 08:41

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