So I was teaching a student high school math and I came upon this problem:

Assuming one has more than oce solution for x, for the equation

$4^{ax} = 8.b^{x}$, find all possible solution of $a, b$.

Point is I can find the answer using trial and error, how can I algebraically solve it?


P.S. Using logarithms isn't allowed, only exponents!

  • $\begingroup$ is this $$4^{ax}=8\cdot b^x$$? $\endgroup$ – Dr. Sonnhard Graubner Mar 18 '17 at 10:41
  • $\begingroup$ @Dr.SonnhardGraubner Yes! $\endgroup$ – Jishan Mar 18 '17 at 10:56

if so we have $$2ax\ln(2)=3\ln(2)+x\ln(b)$$ and then: $$x(2a\ln(2)-\ln(3))=3\ln(2))$$ thus we get $$x=\frac{3\ln(2)}{2a\ln(2)-\ln(3)}$$ for $$b>0,2a\ln(2)-\ln(3)\ne 0$$

  • $\begingroup$ As I said this problem features in the exponent section, using logarithms isn't allowed! $\endgroup$ – Jishan Mar 18 '17 at 10:56
  • $\begingroup$ i think in this method is to find the variable $x$ impossible $\endgroup$ – Dr. Sonnhard Graubner Mar 18 '17 at 11:09
  • $\begingroup$ I thought the same, but the textbook specifically asks the question :( $\endgroup$ – Jishan Mar 18 '17 at 11:11
  • $\begingroup$ maybe there is a typo in your textbook $\endgroup$ – Dr. Sonnhard Graubner Mar 18 '17 at 11:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.