Question 419035
{{{log(7, (3^(x-1)))=2}}}
Solving equations where the variable is in an exponent usually involves the use of logarithms. Logarithms are used because thay have a property, {{{log(a, (p^q)) = q*log(a, (p))}}}, which allows you to move an exponent out in front where we can "get at it" to solve for the variable.<br>
Your equation already has the logarithm in it. So we go straight to using the property to move the exponent:
{{{(x-1)log(7, (3))=2}}}
To solve for x we will start by dividing both sides by that logarithm:
{{{x-1=(2/log(7, (3)))}}}
And last of all, add 1:
{{{x=(2/log(7, (3)))+1}}}
This is an exact expression for the solution. If you need/want a decimal approximation, them use the change of base formula for logarithms, {{{log(a, (p)) = log(b, (p))/log(b, (a))}}}, to convert the base 7 log into an expression of logs whose base your calculator "knows", like base 10 or base e (aka ln). Then use your calculator to find the two logarithms and simplify.