The Unstated Rule of Logarithms

Algebra Level 2

For $a, b, c > 1$ , which of the following is equal to

$\Large \log_{ {\color{#D61F06} a} ^ {\color{#3D99F6} b} } \left( {\color{#20A900}c} \right)?$

\frac{1}{a} \log_b c

\frac{1}{b} \log_a c

b \log_a c

None of the rest

3 solutions

Sandeep Bhardwaj
Nov 22, 2015

To tackle this problem we will use the basic properties of logarithms .

More specifically we will apply the rule $\log_pq=\dfrac{\log q}{\log p}$ . So we can write $\left( \log_{a^b}c\right)$ as $\left( \dfrac{\log c}{\log a^b} \right)$ , which can further be written as $\dfrac{\log c}{b \log a}$ . Applying the above rule in the reverse manner, we get $\dfrac 1b \log_ac .\square$

Moderator note:

Simple standard approach of the rules of logarithms.

I wonder why people hardly mention this. It's a nice application of these rules.

Calvin Lin Staff - 5 years, 6 months ago

May be because they feel that they can derive it easily whenever they require this! [I feel the same]

And the reason supporting this can be: there can exist several such applications, so in that case we ain't going to mention all of them. One of them being like $\log_{a^b}m^n=\dfrac nb \cdot \log_a m$ .

But yeah, I agree that this should be mentioned!

Sandeep Bhardwaj - 5 years, 6 months ago

Hmm.. Fortunately it was already mentioned as a separate property of logarithms in my book :P

Swapnil Das - 5 years, 6 months ago

Yea, it's not in most lists, including our wiki page lol.

Calvin Lin Staff - 5 years, 6 months ago

Shouldn't it be mentioned that a is positive number

Pranav Rao - 5 years, 6 months ago

It is not necessary that $a$ must be positive. If $b$ is an even natural number, then $a$ can take any negative value except -1.

But off course, as the definition of logarithm states the base of the logarithm, which is $a^b$ in this case, must be positive, but not equal to 1. And here it is considered by default.

Sandeep Bhardwaj - 5 years, 6 months ago

@Sandeep Bhardwaj – Thanks. But my doubt was if 'a' was a negative number and b was an even number then we cannot take b outside as 1/b. So I think log(a^b)C =1/ b log(|a|)C. It's like √{x^2}=|x|

Pranav Rao - 5 years, 6 months ago

@Pranav Rao – Yes, you are going the right way. In that case $\log_{a^b}c=\dfrac{\log_{|a|}c}{b}$ .

Sandeep Bhardwaj - 5 years, 6 months ago

Sure, let me also add that $a \neq 1$ just in case.

Calvin Lin Staff - 5 years, 6 months ago

sir, check my solution:

Let given expression is equal to X. then we have to find X. then we can write C= (a^b)^x; hence C= a^(bx); taking log both sides, we have : log(c)=bx log(a); so , x= (1/b)[log(c)/log(a)]; x= (1/b) log (c with base a). hence ans is (b). p.s- sorry my LATEX is not working .

manish kumar singh - 5 years, 6 months ago

Your solution is correct! Great!

And to get your LATEX in work you've to put the stuff inside \ ( ... \ ) without leaving the space between \ and (. For example if you want to write c=(a^b)^x, then using latex it will be $c=(a^b)^x$ .

To learn more about the latex, you can refer to Daniel's note: Beginner LaTex Guide .

Thanks!

Sandeep Bhardwaj - 5 years, 6 months ago

This is seriously an awesome solving strategy...thumbs up and hats off to you!!

Maheerah Shahid - 5 years, 6 months ago

Thanks :)

Have fun solving more and more problems which would definitely enhance your concepts and boost up your confidence!

And always feel free to use my assistance any time, in any regards!

Sandeep Bhardwaj - 5 years, 6 months ago

Hjalmar Orellana Soto
Nov 28, 2015

$log_{a^{b}} c = \frac{1}{blog_{c} a} = \frac{log_{a} c}{b}$

Fætter Guf
Nov 22, 2015

The definition of logarithms states:

" $z$ is the base- $x$ logarithm of $y$ if and only if $x^z = y$ . In typical notation: $\log_x y = z \iff x^z = y.$ "

Applying this to the problem at hand gives us:

$\log_{{a}^{b}} c = z \iff ({a}^{b})^{z} = c$ .

Since $({a}^{b})^{z} = {a}^{bz}$ , we get that ${a}^{bz} = c \iff \log_a c = bz$ or $\frac{1}{b} \log_a c = z$ . We see now that $\log_{{a}^{b}} c = \frac{1}{b} \log_a c$ .

Moderator note:

Great approach of using the definition of logarithms to work out the answer to this question.

extremely sorry, i have written same solution .

manish kumar singh - 5 years, 6 months ago

No problem at all!

Fætter Guf - 5 years, 6 months ago

The Unstated Rule of Logarithms

3 solutions

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