i can't solve this question

if 5^x x 25^2y=1 and 3^5x x 9^y = 1/9 calculate the value of x and y

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- Oct 20th 2009, 12:58 PMdeadmotorExamination questions
i can't solve this question

if 5^x x 25^2y=1 and 3^5x x 9^y = 1/9 calculate the value of x and y - Oct 20th 2009, 01:10 PMpickslides
Can you confrim this to be the system?

$\displaystyle 5^{x} \times 25^{2y}=1 $

$\displaystyle 3^{5x} \times 9^{y} = \frac{1}{9}$ - Oct 20th 2009, 01:12 PMdeadmotor
yeah

how did you do that? - Oct 20th 2009, 01:51 PMpickslides
I made the equations look all pretty using LaTeX. For further information look at

http://www.mathhelpforum.com/math-he...-tutorial.html

I'll kick this off for you.

$\displaystyle 5^{x} \times 25^{2y}=1$ ....(1)

$\displaystyle 3^{5x} \times 9^{y} = \frac{1}{9}$ ....(2)

Let's have a look at (1)

$\displaystyle 5^{x} =25^{-2y}$

$\displaystyle 5^{x} =(5^2)^{-2y}$

$\displaystyle 5^{x} =(5^2)^{-2y}$

$\displaystyle 5^{x} =5^{-4y}$

$\displaystyle x= -4y$

Aim to do the same thing with (2) and you should find a solution. - Oct 21st 2009, 02:42 AMHallsofIvy
here's another way to do that (actually harder- pickslide's method is more fundamental and better):

Given $\displaystyle 5^x25^y= 1$ and $\displaystyle 3^{5x}9^y= 1$, take logarithms of both sides (to any base). log(5^x)+ log(25^y)= log(x) or xlog(25)+ ylog(25)= 0 and since [tex]log(25)= log(5^2)= 2log(5)[tex], x log(5)+ 2y log(5)= 0 and, dividing both sides by log(25), x+ 2y= 0.

Similarly, taking logarithms of both sides of $\displaystyle 3^{5x}9^y= 1$, 5x log(3)+ y log(9)= 5x log(3)+ 2y log(3)= 0 so 5x+ 2y= 0. And it should be pretty obvious that x= 0, y= 0 is the only solution to that.

To see the "Latex" code for $\displaystyle 5^x25^y= 1$ and $\displaystyle 3^{5x}9^y= 1$, click on the formulas.