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
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
$\displaystyle 8^x = 4^{2x+3}$
$\displaystyle (2^3)^x = (2^2)^{2x+3}$
$\displaystyle 2^{3x} = 2^{4x+6}$
$\displaystyle 3x = 4x+6$
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?
$\displaystyle \log_3(2-3x) = \log_9(6x^2 - 19x + 2)$
$\displaystyle \displaystyle \log_3(2-3x) = \frac{\log_3(6x^2 - 19x + 2)}{\log_3{9}}$
$\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
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 $\displaystyle 6x^2- 19x+ 2= 2- 3x$