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I need to find the value of 'x' in the equation $8^x + 2^x = 130$.

As you can see, my final answer at the bottom of the pic is different from what my calculator is saying.

I feel that the step in line 3, which says, $ log (8^x) + log (2^x) = log (130) $ might have something to do with the wrong answer.

Can someone give me feedback on what I worked out?

enter image description here

Alv
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Mars Sojourner
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    You don't have $\log(a+b) = \log ( a) + \log (b)$ but $\log(ab) = \log ( a) + \log (b)$. Try writing $8$ as a power of $2$ – julio_es_sui_glace Jul 18 '23 at 14:05
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    $\log(a+b)\neq \log(a) + \log(b)$ – Eric Jul 18 '23 at 14:05
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    Yes, you can "do the same thing to both sides" and have any solutions to the original also be solutions for the modified... but recall that if you are doing something to a side, you are doing it to the entire side as though that entire side were in parentheses. Yes, if $a=c$ then you have $f(a)=f(c)$. If you have $(a+b)=c$ then $f(a+b)=f(c)$. Most of the time we do not have that $f(a+b)=f(a)+f(b)$, including taking logarithms. – JMoravitz Jul 18 '23 at 14:27
  • @Eric Where do you see me using log(a+b) = log(a) + log(b)? – Mars Sojourner Jul 18 '23 at 14:28
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    "Where do you see me using log(a+b) = log(a) = log(b)?" When you "took the log of both sides" and called this the sum of logs on the left rather than a single log on the left. – JMoravitz Jul 18 '23 at 14:28
  • @JMoravitz So are you saying that 'doing the same thing to both sides' does -not- apply in line 3? – Mars Sojourner Jul 18 '23 at 14:29
  • "Taking the log of both sides" should have taken you from $8^x+2^x=13$ to $\log(8^x+2^x) = \log(13)$. Compare to taking the log of both sides of $c=13$, but where $c$ happens to be $8^x+2^x$. You would have $\log(c)=\log(13)$, and then replacing what $c$ happens to be when written the other way. – JMoravitz Jul 18 '23 at 14:30
  • @JMoravitz Aha! Got it! So my line 3 should have said, log ( 8^x + 2^x ) = log 130. – Mars Sojourner Jul 18 '23 at 14:36
  • You know that $12+13=25$... well consider the action of "adding one" and let's call that function $s$ (s for "successor"). If we were to want to "add one to both sides" this is not $s(12)+s(13)=s(25)$ which would have been $(12+1) + (13+1) = (25+1)$ which you can see is wrong since $27\neq 26$, but rather it is $(12+13)+1 = (25) +1$ – JMoravitz Jul 18 '23 at 14:37
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    One way is to let $y=2^x$ and solve the cubic $y^3+y-130=0$. – IraeVid Jul 18 '23 at 14:38
  • @IraeVid I used your method. Great trick, when we convert everything into a equation. – Mars Sojourner Jul 18 '23 at 16:04

1 Answers1

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Line 3 should have said, $\log ( 8^x + 2^x ) = \log 130$.

It will also help to assume $y = 2^x$.

Then proceed to find the value of $y$, in the equation $y^3 + y -130 = 0$.

David G. Stork
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Mars Sojourner
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