Ciara solved the exponential equation 3x+1 = 15 and her work is shown below. What is the first step she did incorrectly? Step 1: log 3x+1 = log15 Step 2: (x + 1)log 3 = log15 Step 3: log3 = log 15 over x plus 1 Step 4: 0.477121 = 1.176091 over x plus 1 Step 5: 0.477121(x + 1) = 1.176091 Step 6: x + 1 = 1.176091 over 0.477121 Step 7: x + 1 = 2.464975 Step 8: x = 1.464975

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Ciara solved the exponential equation 3x+1 = 15 and her work is shown below. What is the first step she did incorrectly? Step 1: log 3x+1 = log15 Step 2: (x + 1)log 3 = log15 Step 3: log3 = log 15 over x plus 1 Step 4: 0.477121 = 1.176091 over x plus 1 Step 5: 0.477121(x + 1) = 1.176091 Step 6: x + 1 = 1.176091 over 0.477121 Step 7: x + 1 = 2.464975 Step 8: x = 1.464975

Mathematics
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@alexistheking777 what are your thoughts regarding this problem? Do you see which step contains the error?
I think it is either step 2 or 3?

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By the way, if you use \(\LaTeX\) you'll be able to properly post the exponential equation. For example, you can create this: \(3^{x + 1} = 15\)
Oh okay sorry but I think it is either 2 3 or 4 2 is my first choice 4 is my second 3 is my third
Let's go one step at a time. We know the first step is correct because we're allowed to log both sides. What do you think is possibly wrong with the second step?
The fact that x+1 is multiplied to log 3
In order for any step to be wrong, we must be able to identify some mathematical rule that has been violated. Since we're using logs, then we should be able to say which rule of logs was violated for the second step. Do you know what that is (if any)?
No, I'm sorry I have a hard time with logs.
How familiar does the following rule look to you? \(\log(a^b) = b \log(a)\)
pretty familiar
Okay, now suppose \(a = 3\) and \(b = x+1\). How would using that rule of logs I posted above compare with what was done in step 2?
the x+1 can't be multiplied in step 2 because it has to be an exponent i think
So for the rule \(\log(a^b) = b \log(a)\) How would you describe the role of \(b\) on the RIGHT hand side of the equation? Is \(b\) an exponent?
no not on the right side but on the left side ya
\(b\) is not an exponent on the right side. Correct. So in other words, according to the rule, is it possible to re-write \(\log(a^b)\) so that \(b\) is not an exponent?
Yes because it isn't on the right side
Okay, so knowing this, what can we say about step 2?
We can say step 2 is incorrect
Give me like 30 minutes @Hero I'll be back tho.
So for the rule\( log(a^b)=b \log(a)\) : b is allowed to be written as a multiple of log(a) but x+1 isn't allowed to be written as a multiple of log(3) ?
While you're taking a break, you should refer back to the part where I suggested for you to let \(a = 3\) and \(b = x + 1\) for the log rule I already posted.
Hey @Hero I think it can be I think x + 1 can be written as a multiple of log(3)
Yes, I just showed it in the drawing above.
You might have to zoom out to see it all.
I can't see the drawing for some reason. But with that being said it can't be one or two so I think step 4 would be incorrect. I think step 3 is right because she needed to get rid of x+1.
What do you mean "get rid of" x + 1? Why does she "need" to do that?
Because I think you have to do that to simplify but maybe I'm thinking of easier algebra. :(
She "needs" to isolate "x". Dividing both sides by x + 1 is not a step in the direction of isolating x. Dividing both sides by x + 1 isn't getting rid of it. Instead, what it does is limit the domain of x which isn't the goal at all. I guess now you know which step is incorrect.
Thank you can you help me with a couple more if that's alright with you? Your a genius
Post your next question as a new question (separate from this one). Remember anyone can help with a question.
I tagged you in the other one @Hero
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