Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

So, I am unsure what to do next:

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

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

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

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

share|cite|improve this question
The title says 2 variables, the body has only one. Please edit for consistency. – Gerry Myerson Sep 9 '13 at 10:58
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. – Gerry Myerson Sep 9 '13 at 10:59
user1067083, do you think $2^{L\over7}$ looks better than $2^{L/7}$? I don't. – Gerry Myerson Sep 9 '13 at 12:39
up vote 2 down vote accepted

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 $f^{-1}(0)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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