An algebra problem by Gelila Getachew

Algebra Level 3

If $\log_A B = \frac{1}{2}$ and $\log _ C B = \frac{1}{3}$ , what is

$\log_C A ?$

\frac{3}{2}

\frac{1}{6}

\frac{2}{3}

6

2 solutions

Ju Andō
Oct 28, 2015

$B=A^{\frac{1}{2}}\\ B=C^{\frac{1}{3}} \\ \therefore A^{\frac{1}{2}}=C^{\frac{1}{3}}\\ A=C^{\boxed{\frac{2}{3}}}\\$

Gelila Getachew
Nov 28, 2014

logaB.logbC.logcA= (logB/logA).(logC/logB).(logA/logC)=1 therefore, 1/2.1/3.1/k=1 1 / 6k = 1 k = 1 / 6

Can you verify your calculations again? I believe that what you meant is

$\frac{1}{2} \times \frac{1}{3} \times \frac{1}{k} = 1$ , and thus $\frac{1}{6k} = 1$ , instead of $\frac{1}{6} k = 1$ .

Also, note that your answer is currently given as 30. Do you want it to be updated to 6?

Calvin Lin Staff - 6 years, 6 months ago

Yes. Please modify it. I am very sorry. It was a typing error.

Gelila Getachew - 5 years, 11 months ago

If I am not mistaken the value of k should be 1/6 inn order for the multiplication to produce 1

Gelila Getachew - 5 years, 11 months ago

@Gelila Getachew – I'm not sure what you were getting at. I have rephrased the problem. Can you check that it is now correct? If so, please update your solution. Thanks!

Calvin Lin Staff - 5 years, 11 months ago

The phrasing of this problem has been changed. Those who previously answered 3 have been marked correct.

The correct answer to this new version is currently being disputed.

Calvin Lin Staff - 6 years, 6 months ago

An algebra problem by Gelila Getachew

2 solutions

0 pending reports