Question 246776
{{{log(x, (4)) = 1/3}}}
When solving for a variable which is in the base or the argument to a logarithm, you will, as a general rule, rewrite the equation in exponential form. To do this we need to remember that {{{log(a, (p)) = q}}} is equivalent to {{{p = a^q}}}. Using this on your equation we get:
{{{4 = x^(1/3)}}}
Our answer will be "x = some-number" or "some-number = x". These x's do not have a visible exponent. But they do have an exponent. Invisible exponents are 1's. So we want to change {{{x^(1/3)}}} to {{{x^1}}} somehow. This can be done easily if we know how exponents work. (Hint: {{{(a^p)^q = a^(p*q)}}}) We just raise each side to the 3rd power:
{{{(4)^3 = (x^(1/3))^3}}}
which gives us:
{{{64 = x^1}}}
or
{{{64 = x}}}