Answer
The solution is $x=-\dfrac{7}{2}+\dfrac{\sqrt{65}}{2}$
Work Step by Step
$\log_{4}x+\log_{4}(x+7)=1$
Combine the sum on the left side of the equation as the $\log$ of a product:
$\log_{4}x(x+7)=1$
Evaluate the product inside the $\log$ on the left side:
$\log_{4}(x^{2}+7x)=1$
Rewrite in exponential form:
$x^{2}+7x=4^{1}$
$x^{2}+7x=4$
Take $4$ to the left side:
$x^{2}+7x-4=0$
Solve using the quadratic formula, which is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
In this particular case, $a=1$, $b=7$ and $c=-4$.
Substitute the known values into the formula and evaluate:
$x=\dfrac{-7\pm\sqrt{7^{2}-4(1)(-4)}}{2(1)}=\dfrac{-7\pm\sqrt{49+16}}{2}=...$
$...=\dfrac{-7\pm\sqrt{65}}{2}=-\dfrac{7}{2}\pm\dfrac{\sqrt{65}}{2}$
Verify the solutions found by plugging them into the original equation:
$x=-\dfrac{7}{2}-\dfrac{\sqrt{65}}{2}$
$\log_{4}\Big(-\dfrac{7}{2}-\dfrac{\sqrt{65}}{2}\Big)+\log_{4}\Big(-\dfrac{7}{2}-\dfrac{\sqrt{65}}{2}+7\Big)=1$ False
$x=-\dfrac{7}{2}+\dfrac{\sqrt{65}}{2}$
$\log_{4}\Big(-\dfrac{7}{2}+\dfrac{\sqrt{65}}{2}\Big)+\log_{4}\Big(-\dfrac{7}{2}+\dfrac{\sqrt{65}}{2}+7\Big)=1$ True
The solution is $x=-\dfrac{7}{2}+\dfrac{\sqrt{65}}{2}$
