Litzy Downs

2022-10-06

For which values of a does this equation have a solution(s)?

The equation in question is

$${\mathrm{log}}_{5}x\ast ({\mathrm{log}}_{5}(2\ast {\mathrm{log}}_{10}a-x)\ast {\mathrm{log}}_{x}5+1)=2$$

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?

The equation in question is

$${\mathrm{log}}_{5}x\ast ({\mathrm{log}}_{5}(2\ast {\mathrm{log}}_{10}a-x)\ast {\mathrm{log}}_{x}5+1)=2$$

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?

Mario Monroe

Beginner2022-10-07Added 12 answers

Let's start by simplify your equation :

$$\begin{array}{rl}{\mathrm{log}}_{5}x[{\mathrm{log}}_{5}(2{\mathrm{log}}_{10}a-x){\mathrm{log}}_{x}5+1]=2& \iff {\mathrm{log}}_{5}(2{\mathrm{log}}_{10}a-x)+{\mathrm{log}}_{5}x=2\\ & \iff x(2{\mathrm{log}}_{10}a-x)=25\\ & \iff -{x}^{2}+2{\mathrm{log}}_{10}(a)x-25=0\end{array}$$

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 :

$$4{\mathrm{log}}_{10}^{2}a-100\ge 0\iff a\ge {10}^{5}.$$

$$\begin{array}{rl}{\mathrm{log}}_{5}x[{\mathrm{log}}_{5}(2{\mathrm{log}}_{10}a-x){\mathrm{log}}_{x}5+1]=2& \iff {\mathrm{log}}_{5}(2{\mathrm{log}}_{10}a-x)+{\mathrm{log}}_{5}x=2\\ & \iff x(2{\mathrm{log}}_{10}a-x)=25\\ & \iff -{x}^{2}+2{\mathrm{log}}_{10}(a)x-25=0\end{array}$$

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 :

$$4{\mathrm{log}}_{10}^{2}a-100\ge 0\iff a\ge {10}^{5}.$$

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