0
$\begingroup$

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.

$\endgroup$

3 Answers 3

Reset to default
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

Hint:

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

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

$\endgroup$
0
$\begingroup$

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.

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

$|5x+2| = 3x$

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.