Simple logarithms #1

Algebra Level 2

For a , b R + a, b \in \mathbb{R}^+ and > 1 >1 ,

a log a b b log b a = ? \Huge { \color{#D61F06}{a} }^{ \sqrt { \log _{ \color{#D61F06}{a} }{ \color{#3D99F6}{b} } } }-{ \color{#3D99F6}{b} }^{ \sqrt { \log _{ \color{#3D99F6}{b} }{ \color{#D61F06}{a}} } }= \ ?


The answer is 0.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Suppose x = a log a ( b ) . \large x = a^{\sqrt{\log_{a}(b)}}. Then by the change of base rule

log a ( x ) = log a ( b ) log b ( x ) log b ( a ) = 1 log b ( a ) \large \log_{a}(x) = \sqrt{\log_{a}(b)} \Longrightarrow \dfrac{\log_{b}(x)}{\log_{b}(a)} = \dfrac{1}{\sqrt{\log_{b}(a)}}

log b ( x ) = log b ( a ) x = b log b ( a ) . \large \Longrightarrow \log_{b}(x) = \sqrt{\log_{b}(a)} \Longrightarrow x = b^{\sqrt{\log_{b}(a)}}.

The given expression is then equal to x x = 0 . \large x - x = \boxed{0}.

i dont get the part where root loga b became the inverse of root logb a

Aren Nercessian - 5 years, 6 months ago

Log in to reply

In general, the change of base rule gives us that log a ( x ) = log b ( x ) log b ( a ) , \log_{a}(x) = \dfrac{\log_{b}(x)}{\log_{b}(a)}, where a , b , x a,b,x are all positive reals.

If x = b x = b then since log b ( b ) = 1 \log_{b}(b) = 1 we find that log a ( b ) = 1 log b ( a ) . \log_{a}(b) = \dfrac{1}{\log_{b}(a)}.

Now if both a , b a,b are greater than 1 1 then both log a ( b ) \log_{a}(b) and log b ( a ) \log_{b}(a) exceed 0 , 0, which allows us to take the square roots of both sides of this last equation, yielding

log a ( b ) = 1 log b ( a ) . \sqrt{\log_{a}(b)} = \dfrac{1}{\sqrt{\log_{b}(a)}}.

(Note that for this question to make any sense it must be assumed that a , b > 1 , a,b \gt 1, so while this condition is not explicitly stated it seems reasonable to assume it implicitly.)

Brian Charlesworth - 5 years, 6 months ago
Pablo Padilla
Nov 25, 2015

First we will focus on a log a b \large{a}^{\sqrt{\log_{a}{b}}} , which can be expessed as a ln b ln a \large{a}^{\sqrt{\frac{\ln{b}}{\ln{a}}}} according to the change of base rule.

Rationalizing the denominator, we get:

a ln b ln a = a ln b ln a ln a \large{a}^{\frac{\sqrt{\ln{b}}}{\sqrt{\ln{a}}}} =\large{a}^{\frac{\sqrt{\ln{b}\ln{a}}}{\ln{a}}} which can also be expresed as ( a 1 ln a ) ln a ln b \large(\large{a}^{\frac{1}{\ln{a}}})^{\sqrt{\ln{a}\ln{b}}}

Now, by the change of base rule, we have:

ln a = log a a log a e = 1 log a e 1 ln a = log a e \ln{a}=\large{\frac{\log_{a}{a}}{\log_{a}{e}}}=\large{\frac{1}{\log_{a}{e}}} \Rightarrow \frac{1}{\ln{a}}=\log_{a}{e}

Hence ( a 1 ln a ) ln a ln b = ( a log a e ) ln a ln b = e ln a ln b (\large{a}^{\frac{1}{\ln{a}}})^{\sqrt{\ln{a}\ln{b}}}=(\large{a}^{\log_{a}{e}})^{\sqrt{\ln{a}\ln{b}}}= \large{e}^{\sqrt{\ln{a}\ln{b}}}

A similar simplification holds for b log b a \large{b}^{\sqrt{\log_{b}{a}}} , so the original expresion reduces to

e ln a ln b e ln a ln b = 0 \large{e}^{\sqrt{\ln{a}\ln{b}}}-\large{e}^{\sqrt{\ln{a}\ln{b}}}=\boxed{0}

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...