How to Solve for x in a Logarithmic Equation?

[rtwikia]

The question is
\(\displaystyle \log_{x}\left({10}\right)+log(x)=2\)
where (obviously) I have to find x.

I tried changing the base,

$\frac{log(10)}{log(x)}+log(x)=2$
$\frac{1}{log(x)}+log(x)=2$
${log(x)}^{-1}+log(x)=log(100)$​

but I could go no further. Whatever I try, I always got a wrong answer.

By guessing and substitution, I found that the answer should be 10. But is there any formal method to find that out?

 

[Nono713]

Hint: multiply through by $\log(x)$ then substitute $u = \log(x)$, you are left with a quadratic equation in $u$.

 

[Greg]

The question is
\(\displaystyle \log_{x}\left({10}\right)+log(x)=2\)

Is this the correct problem statement? If so, what is the base of $\log(x)$?

 

[rtwikia]

Is this the correct problem statement? If so, what is the base of $\log(x)$?

Should be 10?

 

[Greg]

\(\displaystyle \log_x(10)+\log(10)=2\)

\(\displaystyle \log_x(10)=1\)

Can you continue?

 

[rtwikia]

\(\displaystyle \log_x(10)+\log(10)=2\)

\(\displaystyle \log_x(10)=1\)

Can you continue?

Of course, if it is known that x=10 then I can do that. However I'm not sure that 10 is the answer, so...I'll just try to solve the quadratic(Nerd)

 

[rtwikia]

Hint: multiply through by $\log(x)$ then substitute $u = \log(x)$, you are left with a quadratic equation in $u$.

YES it worked!(Rock)

$\log_{x}\left({10}\right)\cdot\log\left({x}\right)+(\log\left({x}\right))^2=2\cdot\log\left({x}\right)$
$\frac{1}{\log\left({x}\right)}\cdot\log\left({x}\right)+(\log\left({x}\right))^2=2\cdot\log\left({x}\right)$
$1+u^2=2u$
$u^2-2u+1=0$​

$u=\frac{-(-2)\pm\sqrt{4-4}}{2}=1$

$\log\left({x}\right)=1$
$x=10^1=10$

Thank you!:D

 

[Greg]

Yup. I misread the problem. Sorry about that.

 

[rtwikia]

Yup. I misread the problem. Sorry about that.

Never mind. Thanks anyway.:)