do you know this property for logarithms?

Algebra Level 2

x R + , log 2187 ( x ) = a b log 3 ( x ) \forall x \in \mathbb{R}^{+},\quad \log_{\sqrt{2187}} \space (x) = \frac{a}{b} \cdot \log_{3} (x) where a , b a, b are coprime positive integers. Enter 10a + b


The answer is 27.

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

Vilakshan Gupta
Jan 27, 2018

The property used is log b m n a = n m log b a \log_{b^\frac{m}{n}} a = \dfrac{n}{m} \log_{b} a

Since 2187 = 3 7 2 \sqrt{2187} = 3^{\frac{7}{2}} , we have log 3 7 2 ( x ) = 2 7 log 3 ( x ) \log_{3^\frac 72} (x) = \frac{2}{7} \log_3 (x) a = 2 , b = 7 \implies a=2 ,b=7 10 a + b = 27 \implies 10a+b=\boxed{27} .

Correct, Just only a small typo. In the last line, b = 7 b = 7

Guillermo Templado - 3 years, 4 months ago

Log in to reply

Oh...Yes. Thanks.

Vilakshan Gupta - 3 years, 4 months ago
Richard Costen
Jan 28, 2018

Using logarithm property log b a = log b n a n \log_b a=\log_{b^n} a^n log 2187 ( x ) = log 3 7 2 ( x ) = log ( 3 7 2 ) 2 7 ( x 2 7 ) = log 3 ( x 2 7 ) = 2 7 log 3 ( x ) \log_{\sqrt{2187}} (x) =\log_{3^\frac72} (x) = \log_{(3^\frac72)^\frac27} (x^\frac27) = \log_3 (x^\frac27) =\dfrac27 \log_3 (x) Therefore a = 2 , b = 7 a=2, b=7 and 10 a + b = 27 10a+b=\boxed{27}

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...