Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is the equation that I am having troubles with:

$$\large x^{\large\log_{10}5}+5^{\large\log_{10}x}=50$$

So the first thing I do, I logarithm the whole expression with $\log_{10}$.
So I get:

$ {\log_{10} 5} \times {\log_{10} x} + {\log_{10} 5} \times {\log_{10} x} = {\log_{10} 50}$

When I solve this one for $x$, I get that $x = 16$, which is totally incorrect because it is supposed to be $100$. Can anyone tell me what am I doing wrong or show me how to solve this equation?

share|improve this question

3 Answers 3

up vote 11 down vote accepted

Let $y=x^{\large\log_{10}5}$, then $$ \log_{10}y=(\log_{10}5)(\log_{10}x)=\log_{10}5^{\large\log_{10}x}\color{red}{\quad\Rightarrow\quad} y=5^{\large\log_{10}x}. $$ Hence \begin{align} x^{\large\log_{10}5}+5^{\large\log_{10}x}&=50\\ 5^{\large\log_{10}x}+5^{\large\log_{10}x}&=50\\ 2\times5^{\large\log_{10}x}&=50\\ 5^{\large\log_{10}x}&=25\\ 5^{\large\log_{10}x}&=5^2\color{red}{\quad\Rightarrow\quad}\log_{10}x=2\color{red}{\quad\Rightarrow\quad}\large\color{blue}{ x=10^2=100}. \end{align}

share|improve this answer
    
Thanks a lot, this really helped :) –  Kockar May 28 '14 at 11:08
    
You are welcome @Kockar :) –  Tunk-Fey May 28 '14 at 11:11

Your first action was bad, because the identity you assumed ("$\log{a+b}=\log{a}+\log{b}$") simply doesn't exist.

share|improve this answer
    
I have always liked the equation : $\log (1+2+3)=\log 1+\log 2 +\log 3$ –  evil999man May 28 '14 at 12:51
    
@Awesome It is the same as $\log{ab}=\log{a}+\log{b}$ :-) –  peterh May 28 '14 at 12:56

If $x > 100$, then: $LHS \gt 100^{log_{10}^5} + 5^{log_{10}^{100}} = \left(10^{log_{10}^5}\right)^2 + 5^2 = 5^2 + 5^2 = 50 = RHS$,

similarly if $x < 100$, then $LHS < RHS$. Thus $x$ can only be $100$.

share|improve this answer
    
But how do I get that 100 so I can start calculating and comparing? –  Kockar May 28 '14 at 10:50
1  
It can be done without using trial & error like Tunk-Fey's answer –  Anastasiya-Romanova 秀 May 28 '14 at 10:59
    
OK, fine. Whatever... –  Anastasiya-Romanova 秀 May 28 '14 at 11:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.