kin the equation solve the following
16^k=2^2k+6
my classmate tells me the answer is k =3, is this right ?
Muchos gracias
$\displaystyle 16^k=2^{2k}+6$,
may be written:
$\displaystyle 2^{4k}=2^{2k}+6$.
Now put $\displaystyle y=2^{2k}$, and we have a quadratic in $\displaystyle y$:
$\displaystyle y^2=y+6$
or:
$\displaystyle y^2-y=6=0$
which has roots $\displaystyle y=3, \ -2$, but as we are interested in real solutions for k we have $\displaystyle y=-2$ does not give a real $\displaystyle k$, so we are left with $\displaystyle y=3$, then:
$\displaystyle 3=2^{2k}$
take logs:
$\displaystyle \log(3)=2\,k \log(2)$
$\displaystyle k=\log(3)/(2\log(2)) \approx 0.792$
RonL
Hello, Greg!
Solve: .$\displaystyle 16^k\:=\:2^{2k+6}$
We have: .$\displaystyle (2^4)^k\:=\:2^{2x+6}\quad\Rightarrow\quad 2^{4k}\:=\:2^{2k+6}$
Since the bases are equal, the exponents are equal:
. . $\displaystyle 4k\:=\:2k+6\quad\Rightarrow\quad2k\:=\:6\quad\Righ tarrow\quad k \:=\:3$
This is the way I thought of this problem too. Would have been nice if Greg used parentheses. Any way, your answer is off by a factor of 1/2. You went wrong during your substitution; answer should be: ln(3)/(2*ln(2)) or .7925. Obviously this is irrelevant since it's not even the question he's asking.