Question

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

Answer 1

@Hero

Answer 2

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

Answer 3

I think it is either step 2 or 3?

Answer 4

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

Answer 5

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

Answer 6

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?

Answer 7

The fact that x+1 is multiplied to log 3

Answer 8

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)?

Answer 9

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

Answer 10

How familiar does the following rule look to you?

\(\log(a^b) = b \log(a)\)

Answer 11

pretty familiar

Answer 12

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?

Answer 13

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

Answer 14

@Hero

Answer 15

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?

Answer 16

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

Answer 17

\(b\) 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?

Answer 18

Yes because it isn't on the right side

Answer 19

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

Answer 20

We can say step 2 is incorrect

Answer 21

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

Answer 22

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) ?

Answer 23

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.

Answer 24

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

Answer 25

Yes, I just showed it in the drawing above.

Answer 26

You might have to zoom out to see it all.

Answer 27

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.

Answer 28

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

Answer 29

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

Answer 30

@Hero

Answer 31

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.

Answer 32

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

Answer 33

@Hero

Answer 34

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

Answer 35

I tagged you in the other one @Hero