# How do you solve  log_2 x + log_2 x-15 = 4?

Mar 31, 2016

$x \approx 724.08$

#### Explanation:

$1$. Start by adding $15$ to both sides of the equation.

${\log}_{2} x + {\log}_{2} x - 15 = 4$

${\log}_{2} x + {\log}_{2} x = 19$

$2$. Use the logarithmic property, ${\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m} \cdot \textcolor{b l u e}{n}\right) = {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m}\right) + {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{b l u e}{n}\right)$ to simplify the left side of the equation.

${\log}_{2} \left(x \cdot x\right) = 19$

${\log}_{2} \left({x}^{2}\right) = 19$

$3$. Use the logarithmic property, ${\log}_{\textcolor{p u r p \le}{b}} \left({\textcolor{p u r p \le}{b}}^{\textcolor{\mathmr{and} a n \ge}{x}}\right) = \textcolor{\mathmr{and} a n \ge}{x}$, to rewrite the right side of the equation.

${\log}_{2} \left({x}^{2}\right) = {\log}_{2} \left({2}^{19}\right)$

$4$. Since the equation now follows a "$\log = \log$" situation, where the bases are the same on both sides, rewrite the equation without the "$\log$" portion.

${x}^{2} = {2}^{19}$

$5$. Solve for $x$.

sqrt(x^2)=sqrt(2^(19)

$x = {2}^{\frac{19}{2}}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x \approx 724.08 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Note that although when you take the square root of ${2}^{19}$, a negative solution should also be produced, it is not actually a solution because if you substitute $x = - {2}^{\frac{19}{2}}$ back into the original equation, you will notice that you end up taking the logarithm of a negative number, which is not possible.