solve, giving your answers as exact fractions, the simultaneous equations:

8^(x) = 4^(2x+3)

log(base2)y = log(base2)x + 4

('^' meaning 'to the power of')

I am stuck, i dont know how to do it! PLease show your method,

Thanks, azza xxxz

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- Oct 2nd 2010, 09:55 AMazzalogs and exponentials question
solve, giving your answers as exact fractions, the simultaneous equations:

8^(x) = 4^(2x+3)

log(base2)y = log(base2)x + 4

('^' meaning 'to the power of')

I am stuck, i dont know how to do it! PLease show your method,

Thanks, azza xxxz - Oct 2nd 2010, 10:00 AMPlato
- Oct 2nd 2010, 10:05 AMskeeter
$\displaystyle 8^x = 4^{2x+3}$

$\displaystyle (2^3)^x = (2^2)^{2x+3}$

$\displaystyle 2^{3x} = 2^{4x+6}$

$\displaystyle 3x = 4x+6$

I assume the right side of the second equation is $\displaystyle \log_2{x} + 4$ rather than $\displaystyle \log_2(x+4)$ ...

$\displaystyle \log_2{y} = \log_2{x} + 4$

$\displaystyle \log_2{y} = \log_2{x} + \log_2{16}$

$\displaystyle \log_2{y} = \log_2(16x)$

$\displaystyle y = 16x$

can you finish it? - Oct 2nd 2010, 10:09 AMazza
sorry i typed it in incorrecty

8^(y) = 4^(2x+3)

log(base2)y = log(base2)x + 4

so that gives you 3y = 4x +6 which i make into y = 4/3x + 2 and I put it into the y of the other equation. but how do i solve it then? - Oct 2nd 2010, 10:13 AMskeeter
y = 16x ... substitute 16x for y into the first equation.

- Oct 2nd 2010, 10:20 AMazza
thankyou very much!

i am getting x=1/8, y=2, but the textbook answer is x=3/22 y=24/11

is the textbook wrong or am i? - Oct 2nd 2010, 10:28 AMskeeter
- Oct 2nd 2010, 10:31 AMazza
ahh finally i understand, thankyou so much!

- Oct 2nd 2010, 11:30 AMazza
I have ONe other question:

log(base 3)(2-3x) = log(base9)(6x^2 - 19x + 2)

Thankyou x - Oct 2nd 2010, 11:42 AMskeeter
- Oct 2nd 2010, 12:35 PMazza
sorry, i dont see where that leads?

- Oct 2nd 2010, 02:00 PMskeeter
$\displaystyle \log_3(2-3x) = \log_9(6x^2 - 19x + 2)$

$\displaystyle \displaystyle \log_3(2-3x) = \frac{\log_3(6x^2 - 19x + 2)}{\log_3{9}}$

$\displaystyle \displaystyle \log_3(2-3x) = \frac{\log_3(6x^2 - 19x + 2)}{2}

$

$\displaystyle \log_3(2-3x) = \log_3 {\sqrt{6x^2 - 19x + 2}}$

finish it - Oct 3rd 2010, 03:06 AMHallsofIvy
Another way of looking at it: [tex]log_a(x)= y if and only if $\displaystyle x= a^y$ so if $\displaystyle y= log_3(2- 3x)$ then 2- 3x= 3^y[tex] and if $\displaystyle y= log_9(6x^2- 19x+ 2)$ then $\displaystyle 6x^2- 19x+ 2= 9^y$.

But $\displaystyle 9= 3^2$ so $\displaystyle 6x^2- 19x+ 2= (3^2)^y= 3^{2y}= (3^y)^2= 2- 3x$.

Solve $\displaystyle 6x^2- 19x+ 2= 2- 3x$ - Oct 3rd 2010, 04:24 AMskeeter