Question 93753
Solve for x:
{{{Log[x](32) = 5}}} You can rewrite this in exponential form thus:
{{{Log[b](x) = y}}} can be written as: {{{b^y = x}}} in this problem, you'd get:
{{{Log[x](32) = 5}}} can be written as:
{{{x^5 = 32}}} but we can express 32 as {{{2^5}}}, so...
{{{x^5 = 2^5}}} Here we can apply the property of exponential functions:
If {{{a^n = b^n}}} (for n not = 0} then {{{a = b}}}, so...
If {{{x^5 = 2^5}}} then {{{x= 2}}}