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A question asks to solve: $$135000 = \frac{L}{4} + 75 * 2^{\frac{L}{7}}$$

So, I am unsure what to do next:

$$135000 = \frac{L}{4} + 75 * 2^{\frac{L}{7}}$$

$$135000/75 = \frac{L}{4} + 2^{\frac{L}{7}}$$

$$1800 = \frac{L}{4} + 2^{\frac{L}{7}}$$

$$7200 = L + 2^{\frac{L}{7}}$$

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    $\begingroup$ The title says 2 variables, the body has only one. Please edit for consistency. $\endgroup$ – Gerry Myerson Sep 9 '13 at 10:58
  • $\begingroup$ Also, when you divided by 75, you didn't divide the $L/4$ by 75. And when you multiplied by 4, you didn't multiply the $2^{L/7}$ by 4. $\endgroup$ – Gerry Myerson Sep 9 '13 at 10:59
  • $\begingroup$ user1067083, do you think $2^{L\over7}$ looks better than $2^{L/7}$? I don't. $\endgroup$ – Gerry Myerson Sep 9 '13 at 12:39
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The formula $f:\mathbb{R} \rightarrow \mathbb{R}$ defined by $$f(L) = \frac{1}{4}L + 75 \times 2^{L/7}-135000$$ doesn't simplify much.

Computationally, we can see that $f(75.69)<0$ and $f(75.70)>0$. Since $f$ is a continuous and increasing function for $L$, there's a value of $L$ for which $f(0)=0$ and $$75.69 < L < 75.70$$ and no other real solutions to $f(L)=0$.

I suspect you will be stuck only finding an approximate solution via computation. There might be some hope of finding a formula involving the Lambert W function, but the outcome might be as unhelpful as saying the solution is $f^{-1}(0)$.

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