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...

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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...

Algebra
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\(3^{x+1}=15\) ?
x+1 is in the exponent, like that?
yes

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\(3^{x+1}=15\) I would divide by 3 first, \(3^{x}=5\) then, log base 3, \(\log_33^{x}=\log_35\) \(x\log_33=\log_35\) \(x=\log_35\)
that is what I see as the quickest approach to the exact solution.
oh, I will tell you what though....
in other words, \(\Large \log(A^B)~~\ne~~ A~\log(B)\) BUT, \(\Large \log(A^B)~~\ne~~ B~\log(A)\)
in other words, the exponent in log, goes outside the log.
ah okay
the \(\underline{\rm exponent}\)!
ok, so what step is incorrect ?
Step 3?
well, if step 3, right side, says `log(15/3)` and not `(log 15)/3`, then step 3 is incorrect as well.
but how can something be correct if mistake is already made?
well, step 3, if you assume step 2 is given, is right.
OH okay I see now
:)
Thank you!!
@fakeee \(\Large \log(A^B)~~\ne~~ A~\log(B)\) Yes!! \(\Large \log(A^B)~~\ne~~ B~\log(A)\) NNNNNNNNNO
Because \(\Large \log(A^B)~~=~~ B~\log(A)\)
For example: \(\Large \log(10^2)~~=~~ 2~\log(10)=2\) or \(\Large \log(100)~~=2\)
Yeah I meant to say that 2nd eq. is = (not \(\ne\)) apologize... Thanks Loser...(don't mean to offend you xD)

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