Question 250492
{{{8^(3x-2) = 64}}}
The simplest solution is found by recognizing that 64 is a power of 8. So we can rewrite the equation as:
{{{8^(3x-2) = 8^2}}}
Since both sides are results of raising 8 to a power, the exponents must be equal:
{{{3x-2 = 2}}}
Solving this:
{{{3x = 4}}}
{{{x = 4/3}}}<br>
If we don't notice that {{{64 = 8^2}}} we can solve this using logarithms. We can use any base of logarithm. But we should pick a base our calculator can find. So I will use base 10:
{{{log((8^(3x-2))) = log((64))}}}
Now we can use a property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, to move the exponent of the argument out in front. (This is the reason we use logarithms on problems like this: to get the variable out of the exponent.)
{{{(3x-2)log((8)) = log((64))}}}
Now we solve for x. Divide both sides by {{{log((8))}}}:
{{{(3x-2) = log((64))/log((8))}}}
Add 2 to each side:
{{{3x = 2 + log((64))/log((8))}}}
Multiply both sides by 1/3:
{{{x = 2/3 + log((64))/3log((8))}}}
If we enter the expression on the right into our calculator we end up with
x = 1.333333...
which is 4/3 in decimal form.