# Find x for x^(log 5)+5^(log x)=50

1. May 28, 2014

### Govind_Balaji

1. The problem statement, all variables and given/known data

Find $x$ if $x^{log 5}+5^{log x}=50$

2. Relevant equations

3. The attempt at a solution

I tried $x^{log 5}=50-5^{log x}$

and vice versa. Then substituting the value of x in another equation, I end up with irritating results like
$50=50, 0=0, 1=1, x^{log 5}=x^{log 5}$ etc.

Can anyone give me a hint?

2. May 28, 2014

### ehild

Use the identity $a= e^{\log(a)}$

ehild

3. May 28, 2014

### Mentallic

When you end up with a result like that it means that your substitution was a solution to the equality. If you end up with 1=0 or any other variation then your substitution wasn't a solution. The problem you have however is that this hasn't gotten you anywhere closer to finding what x on its own is.

Do what ehild suggested to try solve for x, and then when you have x=... (where it's a number on the right side and hence has no x value in it), try plugging that value in to your equation again and see if you can get it down to 1=1 or 0=0 etc. If you do, you've found a correct value of x.

4. May 28, 2014

### Ray Vickson

Write $x = e^{\, \log \, x}, \; 5 = e^{\, \log \, 5}$.

5. May 28, 2014

### Govind_Balaji

Thanks I got it $x=50$ also $x^{\log 5}=5^{\log x}$.

6. May 28, 2014

### ehild

What is the base of the logarithm?

ehild

7. May 28, 2014

### CAF123

Do you mean to say your answer is that x = 50?

8. May 28, 2014

That's wrong.$x \neq 50$ Try substituting it into the equation and see if it shows correct

9. May 28, 2014

### Govind_Balaji

$\\x^{\log 5}+5^{\log x}=50\\\\$

$\left(10^{\log x }\right )^{\log 5}+\left(10^{\log 5 }\right )^{\log x} =50\\\\$

$10^{\log x \log 5}+10^{\log 5 \log x}=50$

$\text{Since both are same, } x^{log 5}=5^{log x}\\\\$

$5^{log x}+5^{log x}=50\\\\$

$10^{log_{10} x}=50\Rightarrow x=50$

They are common logarithms with base 10. I think you misunderstood it as base=e.

Last edited: May 29, 2014
10. May 29, 2014

### ehild

Correct so far.
That is wrong.

$2(5^{log x})≠10^{log x}$.

From $10^{\log x \log 5}+10^{\log 5 \log x}=50$, $10^{\log x \log 5}=25=5^2$. Take the logarithm of both sides.

Some people use "log" for natural logarithm, others use it for base-10 logarithm. You need to clarify.

ehild

11. May 29, 2014

### Govind_Balaji

$\log x \log 5=2 \log 5\\ \log x=2 \Rightarrow x=100\\$
Am I right?

12. May 29, 2014

### ehild

Yes!!!!! But check it. Plug-in x=100 into the original equation.

ehild

13. May 29, 2014

### tms

Not quite. If
$$\log x\ \log 5 = 2\ \log 5\ \log x$$
then if you divide both sides by $\log x\ \log 5$ you get
$$1 = 2.$$
And if you plug $x = 100$ into the original equation, you get something much larger than 50, or even 100.

The right-hand side of
$$\log x\ \log 5 = 2\ \log 5\ \log x$$
is right, but you need something else on the left-hand side.

Last edited: May 29, 2014
14. May 29, 2014

### ehild

Nobody said that equation was true. The original equation was

$x^{log 5}+5^{log x}=50$ . log means base-10 logarithm. It came out that

$\log x \log 5 = 2\log 5$

Substitute back x=100: log(5)≈0.699, 100log(5)=25, 5log100=25...

ehild

15. May 29, 2014

### Staff: Mentor

Actually as long as you are looking for a symbolic representation of x what is the logarithm base doesn't matter.

16. May 29, 2014

### ehild

Borek,

the problem has been solved correctly (Post #11). And the numerical value of x was needed. tms is wrong.

ehild

17. May 29, 2014

### D H

Staff Emeritus
It certainly does matter. The solution to $x^{\log_b 5} + 5^{\log_b x} = 50$ is x=4 if the base is 2, x=100 if the base is 10, and x=10000 if the base is 100.

It is true that $5^{\log_b x} = x^{\log_b 5}$, regardless of the base.

18. May 29, 2014

### Govind_Balaji

Actually numerical value was needed. My teacher gave me four options a)25 b)50 c)75 d)100.

He told me that simply $\log x$ means $\log_{10} x$.

And $\ln x$ is used for $\log_e x$.

19. May 29, 2014

### tms

It certainly seemed that you said it was correct when the OP asked if it was correct and you said yes. In retrospect you were saying the numerical answer was correct. I took the unadorned log to be a natural log and so got a very different answer.