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anonymous
 one year ago
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
 one year ago
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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Hero
 one year ago
Best ResponseYou've already chosen the best response.3@alexistheking777 what are your thoughts regarding this problem? Do you see which step contains the error?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think it is either step 2 or 3?

Hero
 one year ago
Best ResponseYou've already chosen the best response.3By 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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
 one year ago
Best ResponseYou've already chosen the best response.3Let'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
 one year ago
Best ResponseYou've already chosen the best response.0The fact that x+1 is multiplied to log 3

Hero
 one year ago
Best ResponseYou've already chosen the best response.3In 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
 one year ago
Best ResponseYou've already chosen the best response.0No, I'm sorry I have a hard time with logs.

Hero
 one year ago
Best ResponseYou've already chosen the best response.3How familiar does the following rule look to you? \(\log(a^b) = b \log(a)\)

Hero
 one year ago
Best ResponseYou've already chosen the best response.3Okay, 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
 one year ago
Best ResponseYou've already chosen the best response.0the x+1 can't be multiplied in step 2 because it has to be an exponent i think

Hero
 one year ago
Best ResponseYou've already chosen the best response.3So 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
 one year ago
Best ResponseYou've already chosen the best response.0no not on the right side but on the left side ya

Hero
 one year ago
Best ResponseYou've already chosen the best response.3\(b\) is not an exponent on the right side. Correct. So in other words, according to the rule, is it possible to rewrite \(\log(a^b)\) so that \(b\) is not an exponent?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes because it isn't on the right side

Hero
 one year ago
Best ResponseYou've already chosen the best response.3Okay, so knowing this, what can we say about step 2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We can say step 2 is incorrect

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Give me like 30 minutes @Hero I'll be back tho.

Hero
 one year ago
Best ResponseYou've already chosen the best response.3So 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
 one year ago
Best ResponseYou've already chosen the best response.3While 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
 one year ago
Best ResponseYou've already chosen the best response.0Hey @Hero I think it can be I think x + 1 can be written as a multiple of log(3)

Hero
 one year ago
Best ResponseYou've already chosen the best response.3Yes, I just showed it in the drawing above.

Hero
 one year ago
Best ResponseYou've already chosen the best response.3You might have to zoom out to see it all.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.3What do you mean "get rid of" x + 1? Why does she "need" to do that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because I think you have to do that to simplify but maybe I'm thinking of easier algebra. :(

Hero
 one year ago
Best ResponseYou've already chosen the best response.3She "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
 one year ago
Best ResponseYou've already chosen the best response.0Thank you can you help me with a couple more if that's alright with you? Your a genius

Hero
 one year ago
Best ResponseYou've already chosen the best response.3Post your next question as a new question (separate from this one). Remember anyone can help with a question.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I tagged you in the other one @Hero
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