# How do I solve this cyclic logarithmic equation?

$$5^{\log_2 x}+2x^{\log_5 2}=15$$ 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?

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

## 2 Answers

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.

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

$$x\approx2.898813156$$