Results 1 to 9 of 9

Math Help - Trig equation with sine exponent

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    13

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by Sleight View Post
    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<br />

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

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

    can you take it from here?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2011
    Posts
    13
    Quote Originally Posted by skeeter View Post
    let u = \sin{x}

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

    4 \cdot (4^2)^{u^2} = (4^3)^u<br />

    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} =
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2011
    Posts
    2
    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?
    Last edited by SnowBudgie; March 16th 2011 at 05:21 PM. Reason: Sorry, I am new to Latex.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,817
    Thanks
    316
    Awards
    1
    Quote Originally Posted by SnowBudgie View Post
     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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,307
    Thanks
    1284
    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).
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Mar 2011
    Posts
    13
    I see, SnowBudgie. Thank you for clarifying that.


    Quote Originally Posted by skeeter View Post
    let u = \sin{x}

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

    4 \cdot (4^2)^{u^2} = (4^3)^u<br />

    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1
    Quote Originally Posted by Sleight View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Mar 2011
    Posts
    13
    Quote Originally Posted by Plato View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. exponent equation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 28th 2011, 01:45 PM
  2. Proving Trig Identity: Sine Difference
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: September 4th 2009, 04:02 AM
  3. trig sine angle
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 18th 2008, 08:54 PM
  4. derivative of sine function with exponent x
    Posted in the Calculus Forum
    Replies: 8
    Last Post: October 11th 2008, 09:01 AM
  5. exponent equation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 13th 2007, 10:33 AM

Search Tags


/mathhelpforum @mathhelpforum