Litzy Downs

## Answered question

2022-10-06

For which values of a does this equation have a solution(s)?
The equation in question is
${\mathrm{log}}_{5}x\ast \left({\mathrm{log}}_{5}\left(2\ast {\mathrm{log}}_{10}a-x\right)\ast {\mathrm{log}}_{x}5+1\right)=2$
Tried working this down with the rules of logarithms, got it down to a quadratic equation of $x$ with $a$ as one of its parameters, but I'm sure that's not the right way to do it because ultimately I would get a single value and not an interval.
Tried getting it down to a single logarithm but they get nested because of different bases. Trivial statements such as $a>0$ don't help (me at least).
I'm not really sure I have an idea of what I should do with this, any hints about what should I aim to get?

### Answer & Explanation

Mario Monroe

Beginner2022-10-07Added 12 answers

Let's start by simplify your equation :
$\begin{array}{rl}{\mathrm{log}}_{5}x\left[{\mathrm{log}}_{5}\left(2{\mathrm{log}}_{10}a-x\right){\mathrm{log}}_{x}5+1\right]=2& ⇔{\mathrm{log}}_{5}\left(2{\mathrm{log}}_{10}a-x\right)+{\mathrm{log}}_{5}x=2\\ & ⇔x\left(2{\mathrm{log}}_{10}a-x\right)=25\\ & ⇔-{x}^{2}+2{\mathrm{log}}_{10}\left(a\right)x-25=0\end{array}$
The discriminant of the last equation is :
$\mathrm{\Delta }=4{\mathrm{log}}_{10}^{2}a-100.$
The equation admits real solutions iff $\mathrm{\Delta }\ge 0$ hence :
$4{\mathrm{log}}_{10}^{2}a-100\ge 0⇔a\ge {10}^{5}.$

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?