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

[Govind_Balaji]

Homework Statement

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

Homework Equations

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]

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

ehild

 

[Mentallic]

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 into 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 Statement

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

Homework Equations

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

 

[Govind_Balaji]

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

 

[ehild]

What is the base of the logarithm?

ehild

 

[CAF123]

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

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

 

[adjacent]

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

 

[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.

 

[ehild]

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

\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

 

[Govind_Balaji]

<br /> <br /> \log x \log 5=2 \log 5\\<br /> <br /> \log x=2 \Rightarrow x=100\\<br />
Am I right?

 

[ehild]

<br /> <br /> \log x \log 5=2 \log 5\\<br /> <br /> \log x=2 \Rightarrow x=100\\<br />
Am I right?

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

ehild

 

[tms]

<br /> <br /> \log x \log 5=2 \log 5\\<br /> <br /> \log x=2 \Rightarrow x=100\\<br />
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.

 

[ehild]

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, 100^log(5)=25, 5^log100=25...


ehild

 

[Borek]

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

 

[ehild]

Borek,

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

ehild

 

[D H]

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.

 

[Govind_Balaji]

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 $\log_e x$.

 

[tms]

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, 100^log(5)=25, 5^log100=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.