Question 245374
{{{log(2, (x)) + log(2, (3x+10)) - 3 = 0}}}
To solve logarithmic equations where the variable is in one (or more) arguments of logarithms, we generally start by transforming the equaiton into one of the following forms:<ul><li>log(expression-with-vairable) = other-expression</li><li>log(expression-with-variable) = log(other-expression)</li></ul>
With the non-logarithm term, the 3, the second form will not be easy to achieve. So we will aim for the first form. The first form has a single logarithm. Our equation has two. Somehow we need to combine them. Fortunately we have a property of logarithms which will allow us to combine our two logarithms into one:
{{{log(a, (p)) + log(a, (q)) = log(a, (p*q))}}}
Using this on our equation we get:
{{{log(2, (x*(3x+10))) - 3 = 0}}}
Now we just get that logarithm by itself by adding 3 to both sides of the equation:
{{{log(2, (x*(3x+10))) = 3}}}
We have finally achieved the first form. With this form we solve it by rewriting the equation in exponential form. To do this we need to know that {{{log(a, (p)) = q}}} is equivalent to {{{p = a^q}}}. Using this on our equation we get:
{{{x*(3x+10) = 2^3}}}
which simplifies to:
{{{3x^2 + 10x = 8}}}
This an equation we can solve. It is quadratic so we will get one side equal to zero by subtracting 8 from each side:
{{{3x^2 + 10x - 8 = 0}}}
Now we either factor and use the Zero Product Property or use the Quadratic Formula. This factors fairly easily:
{{{(3x - 2)(x + 4) = 0}}}
According to the Zero Product Property this (or any) product can be zero only if one (or more) factors is zero. So the solution will be:
{{{3x -2 = 0}}} or {{{x + 4 = 0}}}
Solving each of these we get:
{{{3x = 2}}} or {{{x = - 4}}}
{{{x = 2/3}}} or {{{x = - 4}}}<br>
With logarithmic equations we should check our answers, not so much to see if they work but to see if they make an argument to a logarithm zero or negative. Any such answer must be rejected because we cannot allow arguments of logarithms to be zero or negative.<br>
Always use the original equation to check.
{{{log(2, (x)) + log(2, (3x+10)) - 3 = 0}}}
Checking x = 2/3:
{{{log(2, (2/3)) + log(2, (3(2/3)+10)) - 3 = 0}}}
At this point we can see that neither argument will end up negative or zero. So it looks like 3/2 will work. You are welcome to finish the check.<br>
Checking x = -4:
{{{log(2, (-4)) + log(2, (3(-4)+10)) - 3 = 0}}}
At this point we can see that both arguments will become negative. So we can stop here and reject x = -4 as a solution. (Even if only one of the two arguments had become negative or zero, we would still reject the solution.)<br>
So there is just the one solution: x = 2/3.