It seems really hard for me to solve: ${2}^x +3={0.5}^x$

Also consider the case where it is 3.5 instead of 0.5

A three there makes it so hard. I just have no idea how to apply logarithm in case.

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    $\begingroup$ This can be transformed into a quadratic by multiplying by $2^x$. $\endgroup$ – user61527 Jun 6 '14 at 2:15
  • $\begingroup$ $\large 0.5 = {1 \over 2}$ $\endgroup$ – Felix Marin Jun 6 '14 at 2:15
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    $\begingroup$ I got it, thanks $\endgroup$ – most venerable sir Jun 6 '14 at 2:17
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    $\begingroup$ HINT: Consider that this is equivalent to $(2^x)^2 +3\cdot(2^x)-1=0$. As for $3.5^x$, you'll need numerical methods. Math is rarely nice in that it is frequently impossible to "solve" problems analytically. $\endgroup$ – Chris K Jun 6 '14 at 2:55
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    $\begingroup$ That "3" added to the left-hand side of your equation "puts a spanner in the works", since there is no general rule for the logarithm of a sum or difference of two numbers. Having $ \ 0.5^x \ $ on the right-hand side was all right, since the "base" is still a power of 2 . But using bases on the right-hand side which are other than powers of 2 pretty much guarantees that the equation can't be solved algebraically. $\endgroup$ – colormegone Jun 6 '14 at 4:15

We can start by moving everything to the LHS, yielding

$$-\frac{1}{2^x} + 3 + 2^x = 0$$

We multiply through by $2^x$, giving us

$$-1+3(2^x)+(2^x)^2 = 0$$

For clarity, we can substitute $t = 2^x$, and re-arrange to get a nice quadratic equation.


I assume you can take it from here. The roots aren't particularly beautiful, but hey...

For the case of the RHS being $3.5^x$, it would be impossible to solve analytically. Best option here is to use a numerical method such as Newton's Method. This requires a tiny bit of calculus, and will give you an answer that is as accurate as you'd like it to be. Just keep re-iterating the process until you have, say, 20 correct digits.

  • $\begingroup$ Maybe you should demonstrate the Newton's method. $\endgroup$ – most venerable sir Jun 6 '14 at 23:30
  • $\begingroup$ If it was relevant to the problem at hand, and using Newton's Method was a natural part of the equation solution, I probably would. If you're looking for examples, you might want to google "newton's method examples". $\endgroup$ – Alec Jun 7 '14 at 1:34
  • $\begingroup$ It involves more than just algebra. The general Idea I just learned is basically telling how to use derivative to find the equations for tangent lines that would help you find the roots of a equation. $\endgroup$ – most venerable sir Jun 7 '14 at 2:11
  • $\begingroup$ WHAT IS DERIVATIVE? Maybe integral or anti derivative? $\endgroup$ – most venerable sir Jun 7 '14 at 2:11
  • $\begingroup$ I think you need a better understanding of the derivative before taking on Newton's Method. In any case, I can't explain it reasonably in only 500 characters. $\endgroup$ – Alec Jun 7 '14 at 3:02

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