1. ## solve the following

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

2. Hello, Greg!

k 2k+6
16 = 2

my classmate tells me the answer is k =3, is this right ?

I have no idea . . .

What is the question?

3. ## sorry for the unclear question

I cut and paste it from word and it came out badly. I think I have solved problem for now

4. Originally Posted by Greg
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
$16^k=2^{2k}+6$,

may be written:

$2^{4k}=2^{2k}+6$.

Now put $y=2^{2k}$, and we have a quadratic in $y$:

$y^2=y+6$

or:

$y^2-y=6=0$

which has roots $y=3, \ -2$, but as we are interested in real solutions for k we have $y=-2$ does not give a real $k$, so we are left with $y=3$, then:

$3=2^{2k}$

take logs:

$\log(3)=2\,k \log(2)$

$k=\log(3)/(2\log(2)) \approx 0.792$

RonL

5. Hello, Greg!

Solve: . $16^k\:=\:2^{2k+6}$

We have: . $(2^4)^k\:=\:2^{2x+6}\quad\Rightarrow\quad 2^{4k}\:=\:2^{2k+6}$

Since the bases are equal, the exponents are equal:

. . $4k\:=\:2k+6\quad\Rightarrow\quad2k\:=\:6\quad\Righ tarrow\quad k \:=\:3$

6. Originally Posted by CaptainBlack
$16^k=2^{2k}+6$,

may be written:

$2^{4k}=2^{2k}+6$.

Now put $y=2^k$, and we have a quadratic in $y$:

$y^2=y+6$

or:

$y^2-y=6=0$

which has roots $y=3, \ -2$, but as we are interested in real solutions for k we have $y=-2$ does not give a real $k$, so we are left with $y=3$, then:

$3=2^k$

take logs:

$\log(3)=k \log(2)$

$k=\log(3)/\log(2) \approx 1.585$

RonL
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.