a) given that 2^x = 1/[rt2] and 2^y = 4[rt2]
find the exact value of x and the exact value of x
b) given that 4^x = 8^(2-x) find the value of x
c) given that3^x = 9^(y-1) show that x = 2y-2
I'll start you off
$\displaystyle 2^x = \frac 1{\sqrt{2}}$
take log to the base 2 of both sides:
$\displaystyle \Rightarrow \log_2 2^x = \log_2 \left( \frac 1{\sqrt{2}}\right)$
$\displaystyle \Rightarrow x \log_2 2 = \log_2 \left( \frac 1{\sqrt{2}}\right)$
$\displaystyle 2^y = 4 \sqrt {2}$
do the same thing here:
note that
b) given that 4^x = 8^(2-x) find the value of x
$\displaystyle 4 = 2^2$ and $\displaystyle 8 = 2^3$
So we have $\displaystyle \left( 2^2 \right)^x = \left( 2^3 \right)^{2 - x}$
now continue
similarly here, $\displaystyle 3^x = \left( 3^2 \right)^{y - 1}$
c) given that3^x = 9^(y-1) show that x = 2y-2
now continue