Question 711162
Is "x2" {{{x^2}}} or {{{x*2}}}? The problem is much easier if it is x*2. I'll solve it both ways, starting with {{{x^2}}}:
{{{log(10, (4^(2x+3))) = ln(3^(x^2))}}}
On both sides of the equation, the variable is in the exponent of the argument of a log. There is a property of logarithms, {{{log(a, (p^n))  = n*log(a, (p))}}}, which allows us to move such exponents out in front of the log. Using this property on our equation we get:
{{{(2x+3)*log(10, (4)) = (x^2)*ln(3)}}}
On the left side we can use the Distributive Property to multiply:
{{{2x*log(10, (4))+3*log(10, (4)) = (x^2)*ln(3)}}}<br>
With the {{{x^2}}} term this is a quadratic equation. And with the log's in there, the answer is not going to work out to be a nice whole number (or even a fraction). So I am going to use a calculator to find all the logs (and round the log to the nearest 3 decimal places):
{{{2x*0.602+3*0.602 = (x^2)*1.099}}}<br>
which simplifies to:
{{{1.204x+1.806 = 1.099x^2}}}
Subtracting the squared term from each side:
{{{-1.099x^2+1.204x+1.806 = 0}}}
Then using the Quadratic Formula:
{{{x = (-(1.204) +- sqrt((1.204)^2-4(-1.099)(1.806)))/2(-1.099)}}}
Simplifying:
{{{x = (-(1.204) +- sqrt(1.450-4(-1.099)(1.806)))/2(-1.099)}}}
{{{x = (-(1.204) +- sqrt(1.450+7.939))/2(-1.099)}}}
{{{x = (-(1.204) +- sqrt(9.389))/2(-1.099)}}}
{{{x = (-(1.204) +- 3.064)/2(-1.099)}}}
{{{x = (-1.204 +- 3.064)/2.198}}}
which is short for:
{{{x = (-1.204 + 3.064)/2.198}}} or {{{x = (-1.204 - 3.064)/2.198}}}
which simplify as follows:
{{{x = 1.86/2.198}}} or {{{x = (-4.268)/2.198}}}
{{{x = 0.846}}} or {{{x = -1.942}}}
These are decimal approximations for the solutions to your equation if x2 meant {{{x^2}}}<br>
If x2 meant x*2 (or 2x)...
The steps are mostly the same so I'll leave out the commentary except to explain the differences.
{{{log(10, (4^(2x+3))) = ln(3^(2x))}}}
{{{(2x+3)*log(10, (4)) = (2x)*ln(3)}}}
{{{2x*log(10, (4))+3*log(10, (4)) = (2x)*ln(3)}}}
This is not a quadratic equation. With this we want all the x terms on one side and the other terms on the other side. Subtracting the first term of the left side from both sides gives us:
{{{3*log(10, (4)) = (2x)*ln(3)-2x*log(10, (4))}}}
Factoring out x from the terms on the right side:
{{{3*log(10, (4)) = x(2*ln(3)-2*log(10, (4)))}}}
And then dividing by {{{(2*ln(3)-2*log(10, (4)))}}}:
{{{(3*log(10, (4)))/(2*ln(3)-2*log(10, (4))) = x}}}
This is an exact expression for the solution to the equation if x2 meant x*2 or 2x. Without having to deal with a quadratic equation made it much easier to avoid using our calculator (to get decimal approximations).<br>
If you want or need decimal approximations for this solution, you can use the values for the logs in the first solution (when x2 meant {{{x^2}}} and then simplify the left side.