Question

Calculus 1 homework. please help!

Answer 1

\[6\log_{3}X=-3 \] in equivalent exponential form

Answer 2

You have to isolate the log first

Answer 3

@thearcane do I divide by 6 on both sides?

Answer 4

Yep.

Answer 5

okay. what do I do next?

Answer 6

And then you know that if \[\log_{a}X = b \]
So \[a^{b} = X\]

Answer 7

First of all, you can pass the 6 as exponent of the logarithm argument, in this way it becomes:
\[x^6\]
Then, you can apply the template (call it "formula" if you prefer) to transofrm a logarithm in the equivalent exponential form:
\[log_a{b} = c \rightarrow a^c = b \]
In this way, your expression becomes:
\[log_3{x^6} = -3 \rightarrow 3^{-3} = x^6\]
Solving to find the x:
\[x^6 = \left(\frac{1}{3}\right)^3 \]
\[x = \frac{1}{\sqrt{3}}\]

Answer 8

These are my answer choices
\[x ^{6}=-10\]
\[x ^{6}=27\]
\[x ^{3}=\frac{ 1 }{ 27 }\]
\[x ^{3}=10\]

Answer 9

I'm a little puzzled why we ever would see \(x^{6}\) in the solution to this problem. We have a beautiful logarithm simplification, why not use it?

\(6\cdot log_{3}(X) = -3\)

Notice how the Domain is X > 0.

If we do this (log rule): \(log_{3}\left(X^{6}\right) = -3\), we have introduced Domain that wasn't there, before. This is how the solution above was slightly incorrect along the way. \(x^{6} = 1/27\) has more than one solution - all of which must be evaluated for appropriateness.

If instead, we do this (division): \(log_{3}(X) = -3/6 = -1/2\), we're still on the right track, having NOT changed the Domain, and solution is trivial, \(X = 3^{-1/2} = \dfrac{1}{\sqrt{3}}\)

Not really seeing that in the answer choices. Are you SURE you wrote the problem statement correctly?

Answer 10

6⋅log3(X)=−3 I have to write this problem in exponential form

Answer 11

That has been done, above. Not sure what the answer sheet wants.

Answer 12

All of you guys are right! thanks for your help. sorry this site I'm on is really weird