# Thread: Testing solutions in logarithmic equations

1. ## Testing solutions in logarithmic equations

Hello I'm a newbie. I need help with my homework. I have ADD bad. I don't understand this section Could someone please show how to work these problems?

In Ex. 1-8, determine whether each x-value is a solution of the equation.

3. 3e^x+2=75
(b) -2 + ln25
(c) x approx 1.2189

7. ln(x-1)=3.8
(b) x approx 45.7012
(c) x=1 + ln 3.8

2. Originally Posted by asimon2008
Hello I'm a newbie. I need help with my homework. I have ADD bad. I don't understand this section Could someone please show how to work these problems?

In Ex. 1-8, determine whether each x-value is a solution of the equation.

3. 3e^x+2=75
(b) -2 + ln25
(c) x approx 1.2189

7. ln(x-1)=3.8
(b) x approx 45.7012
(c) x=1 + ln 3.8
A) There are two different ways to do this question:
1. Plug in the given results and check if the equation becomes a true statement.
2. Solve the equation for x and check if you've got one of the given results.

B) to #3 (way #2.):
I assume that the original equation reads: 3e^(x+2)=75 . If so:

$3e^{x+2}=75~\implies~e^{x+2}=25~\implies~x+2=\ln(2 5)~\implies~$ $x=-2+\ln(25) \approx 1.218875825...$

$\ln(1 + \ln( 3.8) - 1) = 3.8~\implies~\ln(\ln(3.8)) \neq 3.8$
$\ln(45.7012 - 1) = \ln(44.7012)\approx 3.8000000347...$