# How do you solve log_2 x = log_4 (x+6)?

Using the change of base rule we have that

logx/log2=log(x+6)/(log2^2) => 2*logx=log(x+6)=> x^2=x+6=> x^2-x-6=0=> (x+2)(x-3)=0=> x_1=3,x_2=-2

Only $x = 3$ is acceptable because $x > 0$

Nov 14, 2015

We have to use the properties of logarithms.

#### Explanation:

${\log}_{2} x = {\log}_{4} \left(x + 6\right)$
$\implies \ln \frac{x}{\ln} 2 = \ln \frac{x + 6}{\ln} 4$
$\implies \ln \frac{x}{\ln} 2 = \ln \frac{x + 6}{\ln} \left({2}^{2}\right)$
$\implies \ln \frac{x}{\cancel{\ln 2}} = \ln \frac{x + 6}{2 \cdot \cancel{\ln 2}}$
$\implies 2 \ln x = \ln \left(x + 6\right)$
$\implies \ln \left({x}^{2}\right) = \ln \left(x + 6\right)$
$\implies {x}^{2} = x + 6$
$\implies {x}^{2} - x - 6 = 0$

which is thus reduced to a quadratic equation which can be solved by the quadratic formula to get the two roots: $x = - 2 \mathmr{and} x = 3$

For $x = 3$, the arguments within the logarithms in the original equation come out to be positive, thus $x = 3$ is a valid solution.

But for $x = - 2$, one of the logarithmic argument is $- 2$ , giving rise to ${\log}_{2} \left(- 2\right)$ which is completely meaningless. Thus $x = - 2$ is discarded and cannot be considered as a solution.

Nov 14, 2015

Make both sides have a base of $4$.

#### Explanation:

Change to 4^(log_2(x))=4^(log_4(x+6).

Notice that on the right side, the $4$ and ${\log}_{4}$ will cancel, leaving just $\left(x + 6\right)$.
${4}^{{\log}_{2} \left(x\right)} = x + 6$

Change the $4$ to ${2}^{2}$.
${\left({2}^{2}\right)}^{{\log}_{2} \left(x\right)} = x + 6$
${2}^{2 {\log}_{2} \left(x\right)} = x + 6$

Remember this logarithm rule.
${2}^{{\log}_{2} \left({x}^{2}\right)} = x + 6$

Like before, the $2$ and ${\log}_{2}$ will cancel.
${x}^{2} = x + 6$
${x}^{2} - x - 6 = 0$
$\left(x - 3\right) \left(x + 2\right) = 0$
x=3 or cancel(x=-2

Remember that in ${\log}_{a} \left(b\right) , b > 0$.