Reversing formulas containing exponents

Question

I have these 4 formulas

e = 4 * n^0.6
e = n^3
e = 1.2 * n^3 - 15 * n^2 + 100*n - 140
e = 5 * n^0.75

They take in n and result in e

How on earth would I reverse this to take in e and result in n. I'm not very math savy is another problem.

I mean I get for addition just subtract and for multiplication just divide but even then I don't know what to do when it comes to an equation with multiple parts to it but then you also have exponents and such mixed in I don't know where to begin to reverse these formulas.

Any help or guidance would be most appreciated and this is for a personal project outside my day job, it's not school or job related but it's important to me so I really want to get this down.

Thanks in advance

Answer 1

The problem is relatively simple, and I think a little bit of formal definition will resolve the problem. First, inverse functions. Given a function $f(x)=y$, meaning a function in which we put in $x$ to get $y$, we can define a similar function, $f^{-1}(y)=x$, where we put in $y$ to get $x$. That effectively reverses the function (it’s a little more complex if the original function is not one-to-one, but that doesn’t matter with your 4 functions). To find that function $f^{-1}(y)$, we solve for $y$. For example, if we are given $y=x+1$, we get $x=y-1$.

Now, we introduce some exponent properties. $(a^b)^c=a^{b*c}$. Since raising both sides of an equation to an exponent doesn’t change the validity of the equation, we can use that to solve most of the equations. For example, with the second equation, we can raise both sides to $\frac{1}{3}$, since using the aforementioned property, we get $n$ on the right side.

Your third equation has a very complex answer. I won’t bother giving you the method, but the answer is $4.16667-.284996z+\frac{36.5502}{z}$, where $z=(2\sqrt{(81y^2-16695y+1.3676*10^6)}-18y+1855)^\frac{1}{3}$.

Answer 2

The third equation is a cubic equation in $n$.

Using whole numbers, rewrite it as $$\frac{6 }{5}n^3-15 n^2+100 n-(140+e)=0$$ and follow the steps given in the linked Wikipedia page.

You will quickly find that the equation has only one real root and using the hyperbolic solution you should arrive to $$n=\frac{25}{6}-5 \sqrt{\frac{5}{3}} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{1855-18 e}{375 \sqrt{15}}\right)\right)$$ which does not present any problem with a pocket calculator.

Answer 3

Since you are not so math savvy, I recommend to have a look at this great video of Vi Hart in which she explains that elementary algebra (including fractional exponents) is just $+1$ in a fancy way. For example, $+3 = +1+1+1$. Let's make it fancy. A product is a fancy way of $+1$. For example, $12 = 3 \times 4$ means that $12$ is equal to $+3$ when counting in a plus $4$ kind of way: $12=+4+4+4$. Even more fancy, exponentiation of $n$ is counting in a times $n$ kind of way.

For example, your second equation ($e$ is $n$ to the power $3$) means that $e$ is $+3$ when counting in a times $n$ kind of way: $e=n*n*n$. You already know what $e$ is in terms of $n$ (it's $e=n^3$) and you are interested in knowing what $n$ is in terms of $e$ (spoiler: it's $n=e^{\frac{1}{3}}$). To get to that result, observe that $n$ is one third along the way to $e$ when counting in a times $e$ kind of way:

I highly recommend to make similar plots for equations 1 and 4. Your first equation ($e = 4 \times n^{0.6}$) means that $e$ is $+4$ when counting in a plus $n^{0.6}$ kind of way, so with a similar observation made when solving your second equation, $n^{0.6}$ is $\frac{1}{4}$ when counting in a plus $e$ kind of way. Moreover $0.6 = \frac{3}{5}$, so $\frac{e}{4} = n^{\frac{3}{5}}$. Making again a similar observation as before, you'll find $n = \left(\frac{e}{4}\right)^{\frac{5}{3}}$.

The fourth equation is similar as the first one. Just replace the factor $4$ with a factor $5$ and the exponent $0.6 = \frac{3}{5}$ with $0.75 = \frac{3}{4}$. You'll obtain $n = \left(\frac{e}{5}\right)^{\frac{4}{3}}$.

I hope that helps to better understand the answer of H Huang. Notice that when using $(a^b)^c=a^{b \times c}$ with $c=\frac{1}{b}$, you'll get $a$, as in the solution of your second equation $n=n^{3 \times \frac{1}{3}} = e^{\frac{1}{3}}$.

The third equation is indeed more complicated to explain, but as explained in the answer of Claude Leibovici, you need to find the solution of a cubic equation in $n$.