Question 404463
{{{log((x+1), (27)) = 3}}}
Solving equations where the variable is in the argument or base of a logarithm usually start with transforming the equation into one of the following forms:
log(expression) = other-expression
or 
log(expression) = log(other-expression)<br>
Your equation is already in the first form. So we can proceed to the next step which is to rewrite the equation in exponential form. In general {{{log(a, (p)) = q}}} is equivalent to {{{p = a^q}}}. Using this pattern on your equation we get:
{{{27 = (x+1)^3}}}
This is an equation we can solve. Next we can find the cube root of each side:
{{{root(3, 27) = root(3, (x+1)^3)}}}
which, since {{{27 = 3^3}}}, simplifies to:
3 = x + 1
Subtracting 1 from each side we get:
2 = x<br>
When solving equations like this one, you <i>must</i> check your answer(s). You must ensure that all arguments and bases remain positive. Any "solution" that would make any argument or base of a logarithm zero or negative must be rejected. And these rejected "solutions" can occur <i>even if no mistakes have been made!</i> This is why you must check.<br>
Always use the original equation to check:
{{{log((x+1), (27)) = 3}}}
Checking x = 2:
{{{log(((2)+1), (27)) = 3}}}
which simplifies to:
{{{log(3, (27)) = 3}}}
We can see that the base and argument of the only logarithm are both positive. SO there is no reason to reject this solution. The rest of the check will tell us if we made a mistake. You are welcome to finish the check.<br>
So the only solution to your equation is x = 2.