How to solve for x in exponential eq...

[botley111222]

Plz help...

e^(3x) = 10^(2x) * 2^(1-x)

 

[mmm4444bot]

Raise each side of the equation to the power of 1/x.

Then, taking the natural logarithm on both sides and simplifying leads to the following equation.

(1 - x)/x = [3 - 2*ln(10)]/ln(2)

 

[soroban]

Hello, botley111222!

\(\displaystyle \text{Solve for }x\!:\;\; e^{3x} \;=\; 10^{2x}\cdot 2^{1-x}\)

\(\displaystyle \text{Take logs: }\qquad \ln\left(e^{3x}\right) \;=\;\ln\left(10^{2x}\cdot2^{1-x}\right)\)

. . . . . . . . . . . .\(\displaystyle 3x\ln e \;=\;\ln\left(10^{2x}\right) + \ln\left(2^{1-x}\right)\)

. . . . . . . . . . . . \(\displaystyle 3x\cdot 1 \;=\;2x\ln 10 + (1-x)\ln2\)

. . . . . . . . . . . . . . \(\displaystyle 3x \;=\;2x\ln10 + \ln 2 - x\ln 2\)

. . .\(\displaystyle 3x - 2x\ln10 + x\ln2 \;=\;\ln 2\)

. . . \(\displaystyle x(3 - 2\ln10 + \ln2) \;=\;\ln2\)

. . . . . . . . . . . . . . .\(\displaystyle x \;=\;\frac{\ln 2}{3 - 2\ln10 + \ln2}\)


Edit: corrected one of my favorite typos . . . z instead of x.
.

 

[mmm4444bot]

Since Soroban finished the exercise, I will post my solution, too (which is equivalent).

x = ln(2)/[3 - ln(50)]