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

- anonymous

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- anonymous

@Hero

- Hero

@alexistheking777 what are your thoughts regarding this problem? Do you see which step contains the error?

- anonymous

I think it is either step 2 or 3?

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## More answers

- Hero

By the way, if you use \(\LaTeX\) you'll be able to properly post the exponential equation. For example, you can create this: \(3^{x + 1} = 15\)

- anonymous

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

- Hero

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?

- anonymous

The fact that x+1 is multiplied to log 3

- Hero

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)?

- anonymous

No, I'm sorry I have a hard time with logs.

- Hero

How familiar does the following rule look to you?
\(\log(a^b) = b \log(a)\)

- anonymous

pretty familiar

- Hero

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?

- anonymous

the x+1 can't be multiplied in step 2 because it has to be an exponent i think

- anonymous

@Hero

- Hero

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?

- anonymous

no not on the right side but on the left side ya

- Hero

\(b\) 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?

- anonymous

Yes because it isn't on the right side

- Hero

Okay, so knowing this, what can we say about step 2?

- anonymous

We can say step 2 is incorrect

- anonymous

Give me like 30 minutes @Hero I'll be back tho.

- Hero

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) ?

- Hero

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.

- anonymous

Hey @Hero I think it can be I think x + 1 can be written as a multiple of log(3)

- Hero

Yes, I just showed it in the drawing above.

- Hero

You might have to zoom out to see it all.

- anonymous

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.

- Hero

What do you mean "get rid of" x + 1? Why does she "need" to do that?

- anonymous

Because I think you have to do that to simplify but maybe I'm thinking of easier algebra. :(

- anonymous

@Hero

- Hero

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.

- anonymous

Thank you can you help me with a couple more if that's alright with you? Your a genius

- anonymous

@Hero

- Hero

Post your next question as a new question (separate from this one). Remember anyone can help with a question.

- anonymous

I tagged you in the other one @Hero

- Hero

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