Help with finding an exact solution to the equation. (logs)

[ghi]

Find an exact solution to the equation
7e^3x-2=35

So far I have done;

7e^3x-2=35

3x-2log 7=log35
3xlog7=log35+2
x=log35+2/3log7
x=.9983616527
But when i plug in the answer to the original equation,
7e^{3(.9983616527)-2} ,it comes out to 18.93467897

Help please!

 

[soroban]

Hello, ghi!

Find an exact solution to the equation: .\(\displaystyle 7e^{3x-2} \:=\:35\)

Divide by 7: .\(\displaystyle e^{3x-2} \:=\:5\)

Take logs: .\(\displaystyle \ln\left(e^{3x-2}\right) \:=\:\ln(5)\)

. . . . . \(\displaystyle (3x-2)\underbrace{\ln(e)}_{\uparrow} \:=\:\ln(5)\)
. . . . . . . . . . .\(\displaystyle ^{\text{This is 1}} \)

. . . . . . . . . . \(\displaystyle 3x-2 \:=\:\ln(5) \)

. . . . . . . . . . . . . \(\displaystyle 3x \:=\:\ln(5) + 2\)

. . . . . . . . . . . . . . \(\displaystyle x \:=\:\frac{1}{3}\big[\ln(5) + 2\big]\:\approx\:1.203145971\)

 

[ghi]

thank you so much!