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

  • #1

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?
 

Answers and Replies

  • #2
ehild
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Use the identity ##a= e^{\log(a)}##

ehild
 
  • #3
Mentallic
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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.
 
  • #4
Ray Vickson
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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}##.
 
  • #5
Thanks I got it ##x=50## also ##x^{\log 5}=5^{\log x}##.
 
  • #6
ehild
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What is the base of the logarithm?

ehild
 
  • #7
CAF123
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Thanks I got it ##x=50## also ##x^{\log 5}=5^{\log x}##.
Do you mean to say your answer is that x = 50?
 
  • #8
adjacent
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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
 
  • #9
##\\x^{\log 5}+5^{\log x}=50\\\\##

[itex]\left(10^{\log x }\right )^{\log 5}+\left(10^{\log 5 }\right )^{\log x} =50\\\\[/itex]

[itex]10^{\log x \log 5}+10^{\log 5 \log x}=50[/itex]

##\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:
  • #10
ehild
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##\\x^{\log 5}+5^{\log x}=50\\\\##

[itex]\left(10^{\log x }\right )^{\log 5}+\left(10^{\log 5 }\right )^{\log x} =50\\\\[/itex]

[itex]10^{\log x \log 5}+10^{\log 5 \log x}=50[/itex]


##\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 [itex]10^{\log x \log 5}+10^{\log 5 \log x}=50[/itex], [itex]10^{\log x \log 5}=25=5^2[/itex]. 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
 
  • #11
[itex]

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

\log x=2 \Rightarrow x=100\\
[/itex]
Am I right?
 
  • #12
ehild
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[itex]

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

\log x=2 \Rightarrow x=100\\
[/itex]
Am I right?
Yes!!!!! But check it. Plug-in x=100 into the original equation.

ehild
 
  • #13
tms
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[itex]

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

\log x=2 \Rightarrow x=100\\
[/itex]
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:
  • #14
ehild
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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
 
  • #15
Borek
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Actually as long as you are looking for a symbolic representation of x what is the logarithm base doesn't matter.
 
  • #16
ehild
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Borek,

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

ehild
 
  • #17
D H
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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.
 
  • #18
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##.
 
  • #19
tms
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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.
 

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