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

## Homework Statement

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

## 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?

ehild
Homework Helper
Use the identity ##a= e^{\log(a)}##

ehild

Mentallic
Homework Helper
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?
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.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

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

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

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

ehild
Homework Helper
What is the base of the logarithm?

ehild

CAF123
Gold Member
Thanks I got it ##x=50## also ##x^{\log 5}=5^{\log x}##.
Do you mean to say your answer is that x = 50?

Gold Member
Thanks I got it ##x=50## also ##x^{\log 5}=5^{\log x}##.
That's wrong.##x \neq 50## Try substituting it into the equation and see if it shows correct

##\\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:
ehild
Homework Helper
##\\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\\\\##
Correct so far.
##10^{log_{10} x}=50\Rightarrow x=50##
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.

They are common logarithms with base 10. I think you misunderstood it as base=e.
Some people use "log" for natural logarithm, others use it for base-10 logarithm. You need to clarify.

ehild

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

ehild
Homework Helper
$\log x \log 5=2 \log 5\\ \log x=2 \Rightarrow x=100\\$
Am I right?
Yes!!!!! But check it. Plug-in x=100 into the original equation.

ehild

$\log x \log 5=2 \log 5\\ \log x=2 \Rightarrow x=100\\$
Am I right?
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:
ehild
Homework Helper
Not quite. If
$$\log x\ \log 5 = 2\ \log 5\ \log x$$

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

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

ehild
Homework Helper
Borek,

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

ehild

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

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

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...
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.