Question 267671
{{{log(a, (5)) = -3/4}}}<br>
Eventually we want an equation that says
a = some-number
So we need to get the "a" out of the base of the logarithm. We can so this by rewriting the equation in exponential form:
{{{5 = a^(-3/4)}}}
Now we need to change the exponent on "a" to a 1 (because {{{a^1 = a}}}). The quickest way to do this make clever use of the rules for exponents. I am going to raise both sides of the equation to the -4/3 power. (See what happens to see why.)
{{{(5)^(-4/3) = (a^(-3/4))^(-4/3)}}}
On the right side, the rule is to multiply the exponents. And what do you get when you multiply -3/4 and -4/3? Answer: 1!
{{{5^(-4/3) = a}}}
This may be an acceptable answer. If you want a decimal approximation you can use your calculator on the expression on the left side. Or, for an exact answer with rational denominators...
{{{5^(-4/3) = a}}}
{{{1/5^(4/3) = a}}}
{{{1/root(3, 5^4) = a}}}
{{{(1/root(3, 5^4))(root(3, 5^2)/root(3, 5^2)) = a}}}
{{{root(3, 5^2)/root(3, 5^6) = a}}}
{{{root(3, 5^2)/5^2 = a}}}
{{{root(3, 25)/25 = a}}}