The discussion revolves around solving an equation involving exponents, specifically the expression \([a^{(3x+1)}][b^{(2x-2)}]\). Participants explore how to find the value of \(x\) under different conditions, including whether the expression can equal zero or a positive constant.
Participants generally agree that the expression cannot equal zero if \(a\) and \(b\) are positive real numbers. However, there is no consensus on the best method to solve the equation or how to handle the different terms involved.
Participants have not fully resolved the mathematical steps necessary to isolate \(x\) in the context of the logarithmic equation. The discussion includes assumptions about the positivity of constants \(a\) and \(b\) that may affect the validity of proposed solutions.
tmt said:Hi,
Is there a way to solve such an equation for x?
[a^(3x+1)][b^(2x-2)]Thanks,
Tim
tmt said:In this case, it could be any constant, either 0 or c, it is just because I am curious.
[a^(3x+1)][b^(2x-2)]=0
[a^(3x+1)][b^(2x-2)]=c
I like Serena said:If we assume that $a$ and $b$ are positive real numbers, then it is not possible with 0.
That is because the power of a positive number is never 0.
With a positive constant $c$, I suggest you take the logarithm of both sides.
\begin{array}{}
a^{3x+1}b^{2x-2}&=& c \\
\ln \left( a^{3x+1}b^{2x-2}\right)&=&\ln c \\
(3x+1)\ln a + (2x-2)\ln b&=&\ln c
\end{array}
Perhaps you know how to solve that.