```Question 251005
{{{log((n+2)) + log((8)) = log((n^2 + 7n +10))}}}<br>
Solving equations where the variable is in the argument of one or more logarithms usually involves transforming the equation into one fo the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)<br>
Since your equation has logarithms on both sides of the equation already, we will aim for the second form. We just have to find a way to combine the two logarithms on the left into a single logarithm. Fortunately we have a property of logarithms, {{{log(a, (p)) + log(a, (q)) = log(a, (p*q))}}}, which allows us to do exactly what we want. Using this property on your equation gives us:
{{{log((n+2)*8) = log((n^2 + 7n +10))}}}
which simplifies to:
{{{log((8n+16)) = log((n^2 + 7n +10))}}}<br>
We now have the desired (second) form. The next step uses the idea that if the logarithm of n+2 is the same as the logarithm of {{{n^2 +7b +10}}} then
n+2 must be the same as {{{n^2 +7b +10}}}:
{{{8n+16 = n^2 + 7n +10}}}
The variable is now out of the argument of any logarithms. This is why we use the desired forms. Equations in those forms can be rewritten without logarithms. (BTW: If you use the first form, you rewrite it in exponential form.)<br>
We now have a "normal" quadratic equation to solve. So we'll get one side equal to zero (by subtracting 8n and 16 from ecah side):
{{{0 = n^2 - n - 6}}}
Next we factor or use the Quadratic Formula. This factors pretty easily:
{{{0 = (n-3)(n+2)}}}
By the Zero Product Property this product can be zero only if one of the factors is zero. So:
{{{n-3 = 0}}} or {{{n+2 = 0}}}
Solving these we get:
{{{n = 3}}} or {{{n = -2}}}<br>
With logarithmic equations it is important to check your answers. We must reject any solutions which make an argument of a logarithm negative or zero. Always use the original equation to check your answers.
{{{log((n+2)) + log((8)) = log((n^2 + 7n +10))}}}
Checking x = 3:
{{{log(((3)+2)) + log((8)) = log(((3)^2 + 7(3) +10))}}}
{{{log((5)) + log((8)) = log((9 + 7(3) +10))}}}
{{{log((5)) + log((8)) = log((9 + 21 +10))}}}
{{{log((5)) + log((8)) = log((40))}}}
As you can see, all the arguments of the logarithms are positive. So we have no reason to reject x = 3. To finish the check we can use the property from before to combine the logarithms on the left side:
{{{log((5*8)) = log((40))}}}
{{{log((40)) = log((40))}}} Check!<br>
Checking x = -2:
{{{log(((-2)+2)) + log((8)) = log(((-2)^2 + 7(-2) +10))}}}
{{{log((0)) + log((8)) = log((4 + (-14) +10))}}}
Already we have a problem. We have an argument to a logarithm that is zero. So we must reject this solution. (We only have to find one argument to a logarithm that is zero or negative to reject a solution. It makes no difference that the logarithm on the right ends up with a zero argument. Nor does it make any difference that log(8) is OK.)<br>
So the only solution to your equation is x = 3.
```