Question 202643
If the problem is not: {{{log((3x^2 +2x + 2)) = 0.84509}}}, please repost it<ul><li>Using { }, not [ ], to group things. In Math it is perfectly OK to use [ ], but on Algebra.com it causes some problems.</li><li>Use parentheses, including nested parentheses (parentheses inside of parentheses), to make the meaning of the problem clear.</li></ul><br>
If the problem is {{{log((3x^2 +2x + 2)) = 0.84509}}}, start by rewriting it in exponential form:
{{{3x^2 + 2x + 2 = 10^(0.84509)}}}
{{{10^(0.84509) = 6.9998704114807013}}} Since this is so close to 7 I think you are supposed to use 7. (If not then you will have to use the quadratic formula on the equation with this very long decimal!?)
{{{3x^2 + 2x + 2 = 7}}}
Subtracting 7 from both sides:
{{{3x^2 + 2x - 5 = 0}}}
Factoring:
{{{(3x + 5)(x - 1) = 0}}}
The only way for a product to be zero is if one of the factors is zero. So:
{{{(3x + 5) = 0}}} or {{{(x - 1) = 0}}}
Solving these we get:
{{{x = (-5)/3}}} or {{{x = 1}}}