Question 689700
{{{log(5, (x+3)) = 1}}}
Solving equations where the variable is in the argument of a logarithm usually starts with transforming the equation into one of the following forms:
log(expression) = number
or
log(expression) = log(expression)<br>
Your equation is already in the first form. The next step with this form is to rewrite the equation in exponential form. In general {{{log(a, (p)) = q}}} is equivalent to {{{a^q = p}}}. Using this pattern on your equation:
{{{5^1 = x+3}}}
The left side simplifies:
{{{5 = x+3}}}
Subtracting 3:
{{{2 = x}}}