# Thread: Trig equation with sine exponent

1. ## Trig equation with sine exponent

Okay, this one I don't even know how to start. I really have given this a go for a few days now, but I just am not grasping something. I think my memory has failed me with logarithms.

$4\cdot16^{sin^2(x)}=64^{sin(x)}$

I know 4(16)=64, but that shouldn't fit with this problem. It seems to be more of a distraction. I know that $\log_bx=y$ is the same as $b^y=x$, but I can't quite think of what to do since we have exponents on both sides of the equation.

Any help on getting started would be appreciated.

2. Originally Posted by Sleight
Okay, this one I don't even know how to start. I really have given this a go for a few days now, but I just am not grasping something. I think my memory has failed me with logarithms.

$4\cdot16^{sin^2(x)}=64^{sin(x)}$

I know 4(16)=64, but that shouldn't fit with this problem. It seems to be more of a distraction. I know that $\log_bx=y$ is the same as $b^y=x$, but I can't quite think of what to do since we have exponents on both sides of the equation.

Any help on getting started would be appreciated.
let $u = \sin{x}$

$4 \cdot 16^{u^2} = 64^u$

$4 \cdot (4^2)^{u^2} = (4^3)^u
$

$4 \cdot 4^{2u^2} = 4^{3u}$

$4^{2u^2 + 1} = 4^{3u}$

can you take it from here?

3. Originally Posted by skeeter
let $u = \sin{x}$

$4 \cdot 16^{u^2} = 64^u$

$4 \cdot (4^2)^{u^2} = (4^3)^u
$

$4 \cdot 4^{2u^2} = 4^{3u}$

$4^{2u^2 + 1} = 4^{3u}$

can you take it from here?
Ahh, yes. That is looking familiar now. So I would set the exponents equal to each other and solve, since they both have the same base? (can't remember what that's called)

Before I do that, however, I am curious about how you went from this step to the next, on the left side. Adding one in the exponent is the same as multiplying it by 4?

$4 \cdot 4^{2u^2} =$

$4^{2u^2 + 1} =$

4. If I have

$4^a \cdot 4^b$

That is equivalent to

$4^{a+b}$

In this case,

$a= 1 ~\&~ b = 2u^2$

P.S. Why is the last equation scrunched up and how do I fix it?

5. Originally Posted by SnowBudgie
$a= 1 \& b = 2u^2$

P.S. Why is the last equation scrunched up and how do I fix it?
Are you trying for something like $\displaystyle a = 1 \wedge b = 2u^2$
( a = 1 \wedge b = 2u^2 )
or something like $\displaystyle a = 1 ~\text{and}~b= 2u^2$
( a = 1 ~\text{and}~b= 2u^2 )

Or if you are simply looking for how to make a space, use ~ .

-Dan

6. I would have written it as two separate LaTex's:
[ itex ]a= 1[ /itex ] and [ itex ]b= 2u^2[ /itex ]
(without the spaces, of course).

7. I see, SnowBudgie. Thank you for clarifying that.

Originally Posted by skeeter
let $u = \sin{x}$

$4 \cdot 16^{u^2} = 64^u$

$4 \cdot (4^2)^{u^2} = (4^3)^u
$

$4 \cdot 4^{2u^2} = 4^{3u}$

$4^{2u^2 + 1} = 4^{3u}$

can you take it from here?
Okay, I can follow that to a certain point, but alas I get stuck again.

$2u^2+1~=~3u$

Or,

$2sin^2x+1~=~3sinx$

Not quite sure where to go from here. I don't recognize any familiar identities here.

8. Originally Posted by Sleight
$2u^2+1~=~3u$
Not quite sure where to go from here. I don't recognize any familiar identities here.
Solve $2u^2+1~=~3u$ and get $u=1\text{ or }u=\frac{1}{2}$.

You know that $\sin(x)=u$ so solve for $x$.

9. Originally Posted by Plato
Solve $2u^2+1~=~3u$ and get $u=1\text{ or }u=\frac{1}{2}$.

You know that $\sin(x)=u$ so solve for $x$.
Oh good grief, I didn't recognize the quadratic. So it looks like

$sin(x)~=~\frac{\pi}{2},~\frac{\pi}{6},~\frac{5\pi} {6}$

Thanks everyone.