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## anonymous one year ago 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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1. anonymous

$$3^{x+1}=15$$ ?

2. anonymous

x+1 is in the exponent, like that?

3. anonymous

yes

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

in other words, $$\Large \log(A^B)~~\ne~~ A~\log(B)$$ BUT, $$\Large \log(A^B)~~\ne~~ B~\log(A)$$

8. anonymous

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

9. anonymous

ah okay

10. anonymous

the $$\underline{\rm exponent}$$!

11. anonymous

ok, so what step is incorrect ?

12. anonymous

Step 3?

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

OH okay I see now

17. anonymous

:)

18. anonymous

Thank you!!

19. Loser66

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

20. Loser66

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

21. Loser66

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

22. anonymous

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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