Question

Dylan solved the exponential equation 3x+1 = 15 and his work is shown below. What is the first step he did incorrectly?

Step 1: log3x+1 = log15
Step 2: 3 log(x + 1) = log15
Step 3: log(x + 1) = log 15 over 3
Step 4: log(x + 1) = 0.3920304
Step 5: ln[log(x + 1)] = ln0.3920304
Step 6: x + 1 = −0.936415
Step 7: x = −1.936415

I know it's either step 2 or step 3. This is just all so confusing to me...

Answer 1

\(3^{x+1}=15\) ?

Answer 2

x+1 is in the exponent, like that?

Answer 3

yes

Answer 4

\(3^{x+1}=15\)
I would divide by 3 first,
\(3^{x}=5\)
then, log base 3,
\(\log_33^{x}=\log_35\)
\(x\log_33=\log_35\)
\(x=\log_35\)

Answer 5

that is what I see as the quickest approach to the exact solution.

Answer 6

oh, I will tell you what though....

Answer 7

in other words,

\(\Large \log(A^B)~~\ne~~ A~\log(B)\)

BUT,

\(\Large \log(A^B)~~\ne~~ B~\log(A)\)

Answer 8

in other words, the exponent in log, goes outside the log.

Answer 9

ah okay

Answer 10

the \(\underline{\rm exponent}\)!

Answer 11

ok, so what step is incorrect ?

Answer 12

Step 3?

Answer 13

well, if step 3, right side, says `log(15/3)` and not `(log 15)/3`, then step 3 is incorrect as well.

Answer 14

but how can something be correct if mistake is already made?

Answer 15

well, step 3, if you assume step 2 is given, is right.

Answer 16

OH okay I see now

Answer 17

:)

Answer 18

Thank you!!

Answer 19

@fakeee \(\Large \log(A^B)~~\ne~~ A~\log(B)\) Yes!!
\(\Large \log(A^B)~~\ne~~ B~\log(A)\) NNNNNNNNNO

Answer 20

Because \(\Large \log(A^B)~~=~~ B~\log(A)\)

Answer 21

For example: \(\Large \log(10^2)~~=~~ 2~\log(10)=2\)
or \(\Large \log(100)~~=2\)

Answer 22

Yeah I meant to say that 2nd eq. is = (not \(\ne\)) apologize...
Thanks Loser...(don't mean to offend you xD)