Solving logarithmic equation for x

[Belinda]

I've run into another logarithmic equation I'm not sure how to go about solving. Would appreciate a hint or guidelines:

Solve the equation. (Round your answer to four decimal places.)

e^x − 30e^−x − 1 = 0

Thanks.

 

[daon2]

I've run into another logarithmic equation I'm not sure how to go about solving. Would appreciate a hint or guidelines:

Solve the equation. (Round your answer to four decimal places.)

e^x − 30e^−x − 1 = 0

Thanks.

The natural log function hits every real number, so assume \(\displaystyle x=ln(z)\)

Then you are solving: \(\displaystyle z-\frac{30}{z} -1 = 0\). Then re-substitute to get x.

 

[soroban]

Hello, Belinda!

Solve the equation. .Round your answer to four decimal places.

. . \(\displaystyle e^x - 30e^{-x} - 1 \:=\:0\)

Multiply by \(\displaystyle e^x\!:\;\;e^{2x} - 30 - e^x \:=\:0 \quad\Rightarrow\quad e^{2x} - 3^x - 30 \:=\:0\)

Factor: .\(\displaystyle (e^x+5)(e^x-6) \:=\:0 \)

\(\displaystyle \begin{array}{cccccccccc}e^x+5 \:=\:0 & \Rightarrow & e^x \:=\:\text{-}5 & \Rightarrow & \text{no real sol'n} \\ \\ e^x - 6 \:=\:0 & \Rightarrow & e^x \:=\:6 & \Rightarrow & x \:=\:\ln 6 \end{array}\)


Therefore: .\(\displaystyle x \;=\;1.791759469 \;\approx\;1.7918\)