Question

Find the solution to the equation 64^(4 – x) = 4^(2x)

Answer 1

rewrite 64 as a power of 4

\[(4^3)^{4 - x} = 4^{2x}\]

this can be simplified using power of a power law

\[4^{12 - 3x} = 4^{2x}\]

since the bases are the same equate the powers

12 - 3x = 2x

add 3x to both sides

12 = 5x

just solve for x

Answer 2

So x would be equal to 12/5

Answer 3

correct

Answer 4

Thanks :D

Answer 5

hope it helps

Answer 6

May you help me with finding the solution to the equation of log16 32 = x + 5.

Answer 7

start by looking at the left hand side and simplify the base 16 logs

32 = 2 x 16

so log 32 = log (2 x 16) = log2 + log 16

\[\log _{16} 16 = 1\]

and

\[\log _{16} 2 = \log _{16} \sqrt[4]{16} = \frac{1}{4}\]

so the problem becomes

\[\frac{1}{4} + 1 = x + 5\]

or\[1\frac{1}{4} = x + 5\]

solve for x