solve the equation 10x = e2x-1 using Logarithms

[davehogan]

I have to solve the equation 10^x = e^2x-1 using logs. The answer given is -3.305. I can't get my head round this one at all. If "log10=1" and "In10=2.303", how do you determine the value of "x"? Can anyone explain this for me, please?

 

[srmichael]

I have to solve the equation 10^x = e^2x-1 using logs. The answer given is -3.305. I can't get my head round this one at all. If "log10=1" and "In10=2.303", how do you determine the value of "x"? Can anyone explain this for me, please?

You must take the log of both sides which will allow you to bring down the exponent and then it is algebra from there to solve for x. The thing is, you must lake the log of the same base on both sides of the equal sign. So either one of the following is correct:

\(\displaystyle log(10^x) = log(e^{2x-1})\)

or

\(\displaystyle ln(10^x) = ln(e^{2x-1})\)

P.S. I only showed base 10 and base e in the options above as those buttons are on all calculators, plus it's also easier to take the log of both sides to a base that is equal to one of the bases in the original problem.

 

[JeffM]

I have to solve the equation 10^x = e^2x-1 using logs. The answer given is -3.305. I can't get my head round this one at all. If "log10=1" and "In10=2.303", how do you determine the value of "x"? Can anyone explain this for me, please?

\(\displaystyle f(x) = g(x) \iff log_a\{f(x)\} = log_a\{g(x)\}.\)

The statement above is true only if the logarithms are to the same base on both sides of the equation. What base you use is irrelevant.

\(\displaystyle 10^x = e^{2x-1} \implies log_{10}\left(10^x\right) = log_{10}\left(e^{(2x-1)}\right) = xlog_{10}(10) = log_{10}(e) * (2x - 1)\implies x\left\{1 - 2log_{10}(e)\right\} = -log_{10}(e) \implies\)

\(\displaystyle x = \dfrac{- log_{10}(e)}{1 - 2log_{10}(e)} \approx \dfrac{-0.4343}{1 - 2 * 0.4343} = -3.3052.\) Using logs to base 10.

\(\displaystyle 10^x = e^{2x-1} \implies log_{e}\left(10^x\right) = log_{e}\left(e^{(2x-1)}\right) = xlog_{e}(10) = log_{e}(e) * (2x - 1)\implies x\left\{log_e(10) - 2)\right\} = -1 \implies\)

\(\displaystyle x = \dfrac{- 1}{log_{e}(10) - 2} \approx \dfrac{-1}{2.3026 - 2} = -3.3046.\) Using logs to base e.

 

[lookagain]

I have to solve the equation 10^x = e^2x-1 using logs.

The answer given is x= -3.305.

I can't get my head round this one at all.

If "log10=1" and "In10=2.303" (approximately) ,

how do you determine the value of "x"? You can't attain x = -3.305 with the method below with that approximation. *

Can anyone explain this for me, please?

* You will require a better approximation, such as \(\displaystyle \ ln(10) \ \approx \ 2.3026, \ \ \) so that the desired answer is the

one you gave (when rounded to the third decimal place).

- - - - - - - - - - - - - - - - - - - -- --- - - - - - - - - - - - - - - - - - - - --- - - - - ---


If I were to use the approximation for ln(10) you stated:


\(\displaystyle 10^x \ =\ e^{2x - 1} \ \implies\)


\(\displaystyle ln(10^x) \ = \ ln(e^{2x - 1}) \ \implies\)

\(\displaystyle xln(10) \ = \ 2x - 1 \ \implies\)


\(\displaystyle xln(10) - 2x \ = \ -1 \ \implies\)


\(\displaystyle x[ln(10) - 2] \ = \ -1 \ \implies\)


\(\displaystyle x \ = \ \dfrac{\ -1 \ }{ \ ln(10) - 2 \ } \ \implies \ \)


\(\displaystyle x \ \approx \ \dfrac{ \ -1 \ }{ \ 2.303 - 2 \ } \ \implies \)


\(\displaystyle x \ \approx \ \dfrac{ \ -1\ }{ \ 0.303 \ } \ \implies \)


\(\displaystyle x \ \approx \ -3.300 \ \ne \ -3.305\)

 

[davehogan]

That did it! I've got it at last. Thank you all very much.

 

[HallsofIvy]

I have to solve the equation 10^x = e^2x-1 using logs. The answer given is -3.305. I can't get my head round this one at all. If "log10=1" and "In10=2.303", how do you determine the value of "x"? Can anyone explain this for me, please?

By the way, the abreviation of "natural logarithm" is ln (ell-en), not In! I can only imagine that it is because the "ln" on tiny calculator keys looks like an "I" that that mistake is so common!

 

[davehogan]

Logarithms

I was a bit premature when I said that I had understood the explanations given. I understood the solution with logs to base e, but not with logs to base 10. In JeffM's post of 12.19.2013 he gets to the answer using base 10 too quickly for me to follow. I'm a bit hazy after "xlog₁₀(10)=log₁₀(e) * (2x-1)" All the minus and plus signs have been changed, and the denominator in the last term is virtually "1(-2 * 0.4343). If this is just simple algebra, I will have to brush up on "Equations".

 

[Deleted member 4993]

I suspect you were just staring at his reply instead of working it out independantly!

I have broken up his answer into separate lines - may be that will help.

\(\displaystyle f(x) = g(x) \iff log_a\{f(x)\} = log_a\{g(x)\}.\)

The statement above is true only if the logarithms are to the same base on both sides of the equation. What base you use is irrelevant.

\(\displaystyle 10^x = e^{2x-1} \implies \)

\(\displaystyle log_{10}\left(10^x\right) = log_{10}\left(e^{(2x-1)}\right) \)

\(\displaystyle xlog_{10}(10) = log_{10}(e) * (2x - 1)\)

\(\displaystyle x \ = \ 2x * log_{10}(e) \ - \ 1* log_{10}(e) \)

\(\displaystyle x - 2x * log_{10}(e) = - log_{10}(e)\)

\(\displaystyle x*[1 - 2 log_{10}(e)] \ = - log_{10}(e)\)

\(\displaystyle x \ = \ \dfrac{- log_{10}(e)}{1- 2 log_{10}(e)}\)

\(\displaystyle x = \dfrac{- log_{10}(e)}{1 - 2log_{10}(e)} \approx \dfrac{-0.4343}{1 - 2 * 0.4343} = -3.3052.\) Using logs to base 10.

\(\displaystyle 10^x = e^{2x-1} \implies log_{e}\left(10^x\right) = log_{e}\left(e^{(2x-1)}\right) = xlog_{e}(10) = log_{e}(e) * (2x - 1)\implies x\left\{log_e(10) - 2)\right\} = -1 \implies\)

\(\displaystyle x = \dfrac{- 1}{log_{e}(10) - 2} \approx \dfrac{-1}{2.3026 - 2} = -3.3046.\) Using logs to base e.

 

[JeffM]

I was a bit premature when I said that I had understood the explanations given. I understood the solution with logs to base e, but not with logs to base 10. In JeffM's post of 12.19.2013 he gets to the answer using base 10 too quickly for me to follow. I'm a bit hazy after "xlog₁₀(10)=log₁₀(e) * (2x-1)" All the minus and plus signs have been changed, and the denominator in the last term is virtually "1(-2 * 0.4343). If this is just simple algebra, I will have to brush up on "Equations".

First, I must apologize. Apparently there were some typos in post that may have made it hard to follow. Subhotosh Khan has kindly fixed them.

\(\displaystyle 10^x = e^{2x-1} \implies log_{10}\left(10^x\right) = log_{10}\left(e^{(2x-1)}\right)\) OK to here?

Now one law of logs is \(\displaystyle log_a(b^c) \equiv c * log_a(b).\)

We can apply that law to either side of the equation, but here we do it on both sides to get:

\(\displaystyle log_{10}\left(10^x\right) = log_{10}\left(e^{(2x-1)}\right) \implies x * log_{10}(10) = (2x - 1) * log_{10}(e).\) Any question?

But another law of logs is that \(\displaystyle log_a(a) = 1\). So \(\displaystyle x * log_{10}(10) = x * 1 = x.\)

So \(\displaystyle x * log_{10}(10) = (2x - 1) * log_{10}(e) \implies x = (2x - 1) * log_{10}(e).\)

It is important to remember that the log of a number is just a number. And we solve equations in x by isolating all the terms involving x on one side of the equation and all the other terms on the other side of the equation.

\(\displaystyle x = (2x - 1) * log_{10}(e) = 2x * log_{10}(e) - 1 * log_{10}(e) \implies x - 2x * log_{10}(e) = - log_{10}(e).\) OK so far?

\(\displaystyle x - 2x * log_{10}(e) = - log_{10}(e) \implies x\left\{1 - 2log_{10}(e)\right\} = - log_{10}(e).\)

\(\displaystyle x\left\{1 - 2log_{10}(e)\right\} = - log_{10}(e) \implies x = \dfrac{-log_{10}(e)}{1 - 2log_{10}(e)}.\)

Now use your calculator to find the log to the base 10 of e \(\displaystyle \approx 0.4343.\) Then substitute.

\(\displaystyle x \approx \dfrac{-0.4343}{1 - 2 * 0.4343} \approx -3.3052.\)

 

[Deleted member 4993]

Earlier shown as : [FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]3.3046
11 2/3 minutes you-know-where...[/FONT]

No...No .. You do double time in the corner .... read Jeff's answers carefully....

 

[JeffM]

....I'm "carefully" heading for the corner.....:???:

I'll be there soon enough, denis

 

[davehogan]

Logarithms

I've learnt a very valuable lesson today. "x=1x". I've been wondering where the one comes from when you're showing "1-2" on the left side of the equation. I really have got it this time. Many thanks.

 

[JeffM]

I've learnt a very valuable lesson today. "x=1x". I've been wondering where the one comes from when you're showing "1-2" on the left side of the equation. I really have got it this time. Many thanks.

You are welcome. Sometimes we forget the difficulties that students have in learning something new. I had not thought about the potentially unintuitive nature of the equality x = 1x in many decades, but you reminded me that I drove myself crazy in eighth grade on some problem where I failed to grasp that fact.

 

[davehogan]

Logarithms

And, of course: x=x¹​.