Can you solve this? $15\cdot(3^{x+1}) - 243\cdot(5^{x - 2}) = 0$

Question

This is an exponential functions problem. Please help me with this!

Answer 1

There is no "nice" solution here (EDIT: referred to original question, not the edited one).

If you let $a = \log_3 5$ and $y = 3^x$, you can rewrite the equation as:

$15(y+1) - 243(y^a - 2) = 0 \implies 243y^a - 15y + 471 = 0$

but again there's nothing much you can do with that because $a$ is transcendental (but this form is a little more amenable to a numerical solution).

You can solve it approximately using numerical or graphical methods, and that would give you $x \approx 0.4803$.

EDIT: You have edited the question to be much more amenable to an elementary and exact solution.

If you meant $15\cdot(3^{x+1}) - 243\cdot(5^{x - 2}) = 0$, this is quite easily solvable.

Start by rearranging the equation to $15\cdot(3^{x+1}) = 243\cdot(5^{x - 2})$, then take logs of both sides. It doesn't matter which base you use, as long as you're consistent.

$\displaystyle \log 15 + (x+1)\log 3 = \log 243 + (x-2)\log 5$

Further rearrangement gives:

$\displaystyle x(\log 3 - \log 5) = \log 243 - 2 \log 5 - \log 15 - \log 3$

and the final solution is:

$\displaystyle x = \frac{\log 243 - 2 \log 5 - \log 15 - \log 3}{(\log 3 - \log 5)}$

I'll leave the final simplifications using basic log rules to you. You should get a very simple final answer.