How to isolate X in ${A * B ^X = C * D ^ X}$

Question

${A * B ^X = C * D ^ X}$

The idea is to find in how much time (X) a small (A) investment with a good tax (B) beats a big investment (C) with a bad tax (D). All values are nonzero and positive.

Answer 1

$$\ln\left(A\cdot B^X\right)=\ln\left(C\cdot D^X\right)$$

$$\Leftrightarrow\ln\left(A\right)+\ln\left(B^X\right)=\ln\left(C\right)+\ln\left(D^X\right)$$

$$\Leftrightarrow\ln\left(A\right)+X\ln\left(B\right)=\ln\left(C\right)+X\ln\left(D\right)$$

$$\Leftrightarrow X\ln\left(B\right)-X\ln\left(D\right)=\ln\left(C\right)-\ln\left(A\right)$$

$$\Leftrightarrow X\left(\ln\left(B\right)-\ln\left(D\right)\right)=\ln\left(C\right)-\ln\left(A\right)$$

$$\Leftrightarrow X=\frac{\ln\left(C\right)-\ln\left(A\right)}{\ln\left(B\right)-\ln\left(D\right)}$$

Answer 2

Take the natural logarithm of both sides and use the following laws:

• $\log(xy)=\log(x)+\log(y);$
• $\log(x^y)=y\log(x).$

Rearrangement should then give you what you desire.

Answer 3

Let us suppose wlog $a,b,c,d>0$ (it seems obvious from what you said). Suppose moreover $b\neq d$.

$$ ab^x=cd^x\Longleftrightarrow \frac ac=\frac{d^x}{b^x}=\left(\frac db\right)^x $$ Now let us take the logarithm in base $d/b$ which is positive and different from $1$: you get $$ \log_{\frac db}\left(\frac ac\right)=\log_{\frac db}\left(\frac db\right)^x=x\;\;. $$ If otherwise $b=d$, it's clear that $ab^x=cd^x\Longleftrightarrow a=c$ and in this case the equation holds for every $x\in\Bbb R$.