Logarithmic Equation with absolute value

Question

I am trying to solve the following logarithmic equation:

$$\log _{x^{2}}\left | 5x+2 \right |-\frac{1}{2}=\log _{x^{4}}9$$

Surprisingly, the absolute value is not the problem, I have created this set of equations instead of the original one:

$$\left\{\begin{matrix} \log _{x^{2}}(5x+2)-\frac{1}{2}&=\log _{x^{4}}9 & x\geq -\frac{2}{5}\\ \log _{x^{2}}(-5x-2)-\frac{1}{2}&=\log _{x^{4}}9 & x< -\frac{2}{5}\\ \end{matrix}\right.$$

It is the logarithmic equations I struggle with...

Any assistance will be most appreciated. Thank you.

Answer 1

Well, $a=\log_{x^4} 9$ is the power you have to raise $x^4$ to, in order to get $9$. So $(x^4)^a=9$. So $(x^2)^{a} = 3.$ So $a=\log_{x^2} 3$.

Then

$$\log_{x^2}|5x+2|-\log_{x^2}3 =\frac{1}{2}$$

$$(x^2)^{1/2} = \frac{|5x+2|}{3}$$

$$3|x| =|5x+2|$$

Etc.

Answer 2

Hint:

$$\log_{x^4}y=2\log_{x^4}y\;\forall y>0. $$

Rewrite the equation as an equation of the form $\;\log_{x^2}A(x)=\frac12$.

Answer 3

Very strange to represent logarithms with a variable as the base...

We can fix that.

Using.. $\log_a x = \frac {\log x}{\log a}$

we can say:

$\frac {\log |5x+2|}{2\log x^2} + \frac 12 = \frac {\log 9}{\log x^4}$

Then simplify using our log rules.

$\frac {\log |5x+2|}{2\log x} - \frac 12 = \frac {2\log 3}{4\log x}\\ \log |5x+2|- \log x = \log 3\\ \log x|5x+2| = \log 3x$

$|5x+2| = 3x$