Thread: Balancing an Equation

1. Balancing an Equation

So I have an equation that can be represented as follows:

Y=A*B^X

I need to isolate the value of X, but I don't know how to do it and can't seem to refine my online searched well enough to get something that gives me the answer. Any help would be appreciated.

2. Originally Posted by Fooglmog
So I have an equation that can be represented as follows:

Y=A*B^X

I need to isolate the value of X, but I don't know how to do it and can't seem to refine my online searched well enough to get something that gives me the answer. Any help would be appreciated.
$\displaystyle X = \log_B\left( \frac{Y}{A}\right)$.

3. Originally Posted by Fooglmog
So I have an equation that can be represented as follows:

Y=A*B^X

I need to isolate the value of X, but I don't know how to do it and can't seem to refine my online searched well enough to get something that gives me the answer. Any help would be appreciated.
"unpeel" x doing the opposite (inverse) of everything that is done to it. Here, if we were given X and asked to find Y, we would (1) compute B^X, (2) multiply by A.

Do the opposite of each of those, in the opposite order. Since "multiply by A" would be the last thing done in calculating Y, the first thing we do in solving for X is "divide by A":

Divide both sides of the equation by A to get $\frac{Y}{A}= B^x$. Now, to get rid of the " $B^X$" we need to do the opposite of that, and the opposite of an exponential is a logarithm. Specifically, $log_B(x)$ is defined as the inverse function to $B^x$: $log_B(B^x)= x$ and $B^{log_B(x)}= x$.

Taking the logarithm, base B, of both sides of $B^X= \frac{Y}{A}$ gives $X= log_B(\frac{Y}{A})$.

One nice thing about logarithms is that logarithms to all bases are equivalent. While the form that both mr fantastic and I give is probably the nicest, if you have specific values for A and B, you may find that your calculator does not have a "base B" key for your value of B. In that case, you can use common logarithm, log, (base 10) or natural logarithm, ln, (base e):
$log(B^X)= X log(B)= log(Y/A)$ so that $X= log(Y/A)/log(B)$
or
$ln(B^X)= Xln(B)= ln(Y/A)$ so that $X= ln(Y/A)/ln(B)$

You could also use the fact that log(u/v)= log(u)- log(v) to write these as
$X= log_B(Y)- log_B(A)$
$X= (log(Y)- log(A)/log(B)$
or
$X= (ln(Y)- ln(A))/ln(B)$

4. Beautiful. Thank you both.