```Question 633354
{{{log(x, (32))=5/3 }}}
You made a great first step with
{{{x^((5/3)) = 2^5}}}
Not only did you get the variable out of the logarithm, but you recognized that {{{32 = 2^5}}} which turns out to be a very helpful.<br>
But it's not a good idea to go back to log form. That just puts the x back in the logarithm which is not where we want it. Instead, what we want is to change the exponent on x to a 1. If we can do that, we'll be finished since {{{x^1 = x}}}!<br>
So how do we change an exponent? Well there are several rules for exponents that tell us: "When we do __________, we change the exponents in this way.". If we use the multiplication rule (add the exponents) to get an exponent of 1 (by multiplying by {{{x^((-2/3))}}} we could get the exponent of 1 on the left but then we would have {{{x^((-2/3))}}} on the the right side. If we use the division rule (subtract the exponents) we have a similar problem, We can get the exponent we want on the left but we end up with an x on the right side. The rule that will help us is the power of a power rule (multiply the exponents). If we raise both sides to the right power, we can get an exponent of 1 on the left and there will not be any x's on the right.<br>
So what power to we raise {{{x^((5/3))}}} to to get an exponent of 1? Knowing that we will be multiplying this exponent by 5/3 and knowing that multiplying reciprocals <i>always</i> results in 1, the right power to use is the reciprocal of 5/3: 3/5.
{{{(x^((5/3)))^((3/5)) = (2^5)^((3/5))}}}
On the left, we get the 1 we wanted. On the right, after we multiply 5 and 3/5, we get:
{{{x = 2^3}}}
which simplifies to:
{{{x = 8}}}<br>
And, as usual with problems where the variable was in the argument or base of a logarithm, we must check the solution to ensure that all arguments and bases of all logarithms remain valid (positive for both and bases cannot be a 1 either). If any argument or base ends up being invalid we must reject that "solution", even if it's the only one we found!<br>
Use the original equation to check:
{{{log(x, (32))=5/3 }}}
Checking x = 8:
{{{log((8), (32))=5/3 }}}
We can quickly see that the base is positive (and not a 1) and the argument is positive. So our solution checks!```