Logaritm and its properties

Algebra Level 3

x x and y y are real numbers satisfying the equations below.

{ log 10 ( 2 x ) + log 10 y = 2 log 10 x 2 log 10 ( 2 y ) = 4 \begin{cases} \log_{10}(2x) + \log_{10} y = 2 \\ \log_{10} x^2 - \log_{10} (2y) = 4 \end{cases}

If x + y = m n x + y = \dfrac mn , where m m and n n are positive coprime integers, find m 3 n 6 m-3n^6 .


The answer is 9.

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3 solutions

James Watson
Aug 20, 2020

Notation: log x = log 10 x \log x = \log_{10} x

We can use log properties to make both equations look nicer. Let's start with the top one: log 2 + log y = 2 log 2 x y = log 100 2 x y = 100 \begin{aligned} \log 2 + \log y &= 2 \\ \Longrightarrow \log 2xy &= \log 100 \\ \Longrightarrow 2xy &= 100 \end{aligned}

Now for the second one: log x 2 log 2 y = 4 log x 2 2 y = log 1 0 4 x 2 2 y = 1 0 4 \begin{aligned} \log x^2 - \log 2y &= 4 \\ \Longrightarrow \log \frac{x^2}{2y} &= \log 10^4 \\ \Longrightarrow \frac{x^2}{2y}&=10^4 \end{aligned}

We can rearrange the top one for 2 y 2y : 2 x y = 100 2 y = 100 x \begin{aligned} 2xy &= 100 \\ \Longrightarrow 2y &= \frac{100}{x} \end{aligned}

We can plug that into the second equation and solve for x x : x 2 2 y = 1 0 4 x 2 100 x = 1 0 4 x 3 100 = 1 0 4 x 3 = 1 0 6 x = 100 \begin{aligned} \frac{x^2}{\blue{2y}} &=10^4 \\ \Longrightarrow \frac{x^2}{\blue{\frac{100}{x}}} &= 10^4 \\ \Longrightarrow \frac{x^3}{100} &= 10^4 \\ \Longrightarrow x^3 &= 10^6 \\ \Longrightarrow x &= \boxed{100} \end{aligned}

We can use this to solve for y y : 2 y = 100 x 2 y = 100 100 = 1 y = 1 2 \begin{aligned} 2y &= \frac{100}{\blue{x}} \\ \Longrightarrow 2y &= \frac{100}{\blue{100}} = 1 \\ \Longrightarrow y &= \boxed{\frac{1}{2}} \end{aligned}

Now that we have x x and y y , we can see that x + y = 201 2 x + y = \cfrac{\blue{201}}{\orange{2}} where m = 201 \blue{m=201} and n = 2 \orange{n=2} . Now we can use this to find our answer: m 3 n 6 = 201 3 ( 2 ) 6 = 201 192 = 9 \blue{m} - 3\orange{n}^6 = \blue{201} - 3(\orange{2})^6 = 201 - 192 = \green{\boxed{9}}

Chew-Seong Cheong
Aug 20, 2020

{ log 10 ( 2 x ) + log 10 y = log 10 2 + log 10 x + log 10 y = 2 . . . ( 1 ) log 10 x 2 log 10 ( 2 y ) = 2 log 10 x log 10 2 log 10 y = 4 . . . ( 2 ) \begin{cases} \log_{10}(2x) + \log_{10} y = \log_{10} 2 + \log_{10} x + \log_{10} y = 2 & ...(1) \\ \log_{10} x^2 - \log_{10} (2y) = 2 \log_{10} x - \log_{10} 2 - \log_{10} y = 4 &...(2) \end{cases}

From ( 1 ) + ( 2 ) (1)+(2) :

3 log 10 x = 6 log 10 x = 2 x = 1 0 2 = 100 \begin{aligned} 3\log_{10} x & = 6 \\ \log_{10} x & = 2 \\ \implies x & = 10^2 = 100 \end{aligned}

From ( 1 ) (1) :

log 10 2 + 2 + log 10 y = 2 log 10 y = log 10 2 y = 1 2 \begin{aligned} \log_{10} 2 + 2 + \log_{10} y & = 2 \\ \log_{10} y & = - \log_{10} 2 \\ \implies y & = \frac 12 \end{aligned}

Therefore x + y = 100 + 1 2 = 201 2 x+y = 100 + \dfrac 12 = \dfrac {201}2 . Then m 3 n 6 = 201 3 ( 2 6 ) = 9 m - 3n^6 = 201 - 3(2^6) = \boxed 9 .

Priyanshu Singh
Aug 20, 2020

log 2x+log y=2
2xy=100 ....... (1)
and log x square-log 2y=4
x square/2y=10000 .......(2)
from 1 and 2
x cube=10 power 6
y=1/2
x+y=100+1/2=201/2
m=201 and n=2
m-3n power 6=201-3(2)^6=201-192=9




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