Solve for x
2e^(3x) = 4e^(5x)
Hello, cm3pyro!
Solve for $\displaystyle x\!:\;\;2e^{3x} \:= \:4e^{5x}$
We have: .$\displaystyle 4e^{5x} \:=\:2e^{3x}$
Divide by $\displaystyle 4e^{3x}\!:\;\;e^{2x} \:=\:\frac{1}{2}$
Take logs: .$\displaystyle \ln(e^{2x}) \:=\:\ln\left(\frac{1}{2}\right)$
Then: .$\displaystyle 2x\!\cdot\!\ln(e) \:=\:\ln(2^{-1}) \quad\Rightarrow\quad 2x\cdot1 \:=\:-\ln(2) $
Therefore: . $\displaystyle x \;=\;-\frac{1}{2}\ln(2)$
Hello again, cm3pyro!
Write it out!I dont understand what you did here.
Divide by $\displaystyle 4e^{3x}\!:\;\;e^{2x} \:= \:\frac{1}{2}$
We had: .$\displaystyle 4e^{5x} \;=\;2e^{3x}$
Divide by $\displaystyle 4e^{3x}\!:\;\;\frac{4e^{5x}}{4e^{3x}} \;=\;\frac{2e^{3x}}{4e^{3x}} $
. . and that reduces to: .$\displaystyle e^{2x} \:=\:\frac{1}{2}\quad\hdots\;\text{ Get it?}$