# Comparing indices of different bases

Given $a,b, c, d \in \mathbb{R},$ if $2^{a}3^{b}= 2^{c}3^{d},$ can we conclude $a=c, b=d \ ?$

I think we could not do so, for eg, let $a=1,b=0,d=c$, then $2^{1-c}= 3^{c}$ holds for exactly one $c\neq 1,0.$

So under what conditions can we conclude $a=c, b=d \ ?$

Your counterexample is valid, so your conclusion is correct. (But you can easily give $c$ explicitly, so you might as well: $2^1=2^c3^c=6^c$, therefore $c=\log 2/\log 6$.)

If $a,b,c,d$ are restricted to be rational, then we can conclude that $a=c$ and $b=d$. This is because we have $2^r=3^s$ for rationals $r,s$ (just putting $r=a-c$ and $s=d-b$), so if $s \ne 0$ we get $\log_2 3 = r/s$. But we know that $\log_2 3$ is irrational (see the first paragraph of this answer for an easy proof), so $r=s=0$, which means $a=c$ and $b=d$.

We can conclude that $a=c, b=d$ where $a,b,c,d \in \mathbb{R}$ when $a,b,c,d \in \mathbb{N}$ since in this case this becomes the prime number representation of integers which is unique (by the fundamental theorem of arithmetic). By the same token the condition $a=c, b=d$ can be extended over both $\mathbb{Z}$ and $\mathbb{Q}$ (i.e integers and rationals)

In the more general case we have:

$2^{a}3^{b}= 2^{c}3^{d}$

$a \times ln(2) + b \times ln(3) = c \times ln(2) + d \times ln(3)$

$ln(2) \times (a-c) = ln(3) \times (d-b)$

$\frac{a-c}{d-b} = \frac{ln(3)}{ln(2)}$

So for any three choices of the parameters, one can have a solution of the remaining parameter (when $d \ne b$ and $a,b,c,d \in \mathbb{R}$), thus infinite solutions in $\mathbb{R}$