3
$\begingroup$

$$5^{\log_2 x}+2x^{\log_5 2}=15$$ I have also noticed, that logarithmic terms are cyclic and tried to express one as y to make it easier, but still had no luck solving it. Any help?

$\endgroup$
  • $\begingroup$ i think a numerical method will help you $\endgroup$ – Dr. Sonnhard Graubner Mar 15 '17 at 13:59
  • $\begingroup$ I don't think I know this topic yet. How about solving it using only basic properties of logarithms? $\endgroup$ – John Robbers Mar 15 '17 at 14:03
  • $\begingroup$ i think this doesn't work $\endgroup$ – Dr. Sonnhard Graubner Mar 15 '17 at 14:04
  • $\begingroup$ Are you sure about the equation ? Check for typos. $\endgroup$ – Yves Daoust Mar 15 '17 at 14:24
2
$\begingroup$

You can write the equation as

$$x^{\log_25}+2x^{\log_52}=15,$$or with $t=x^{\log_25}$,

$$t+2t^{\log_5^22}=15.$$

There is a single root near $x=2.8988$, with no closed-form expression.

$\endgroup$
2
$\begingroup$

Note that

$$a^{\log_cb}=b^{\log_ca}$$ SO $$5^{\log_2 x}+2x^{\log_5 2}=x^{\log_2 5}+2x^{\log_5 2}=15$$

Let us call $y=\log_25$ So$$x^y+2x^{1/y}=15$$

which has a solution at

$$x\approx2.898813156$$

$\endgroup$

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.