Question 192060
If "a" is a real number and {{{a^x=b}}}, we can isolate "x" by applying the logarithm base 10 to both sides to get {{{log(10,(a^x))=log(10,(b))}}}. 



From there, pull down the exponent "x" to get {{{x*log(10,(a))=log(10,(b))}}}




Now divide both sides by {{{log(10,(a))}}} to get {{{x=log(10,(b))/log(10,(a))}}}



Combine the terms on the right side (using the change of base formula) to get {{{x=log(a,(b))}}}



So after isolating "x", we get {{{x=log(a,(b))}}}



Note: it turns out that {{{b^y=x}}} <====> {{{log(b,(x))=y}}}. In other words, we can convert to and from exponential and logarithmic form.