Results 1 to 7 of 7

Math Help - a(b)^x = a(b)^x ( getting x value )

  1. #1
    Newbie jessyc's Avatar
    Joined
    Jun 2007
    Posts
    18

    a(b)^x = a(b)^x ( getting x value )

    i have a question here and i'm not sure how to tackle this sucker...

    Same deposits $800 into an investment fund that erns 8% per year, compounded annually. Jessica deposits $1000 into an investment fund that earns 6% per year, compounded annually. When will their investments be equal( 3 places after the decimal)?

    thats the question and heres what i wrote down...

    Sam
    a= 800
    b= 8% = 1.08
    c= (null)
    x= (missing value)

    Jessica
    a= 1000
    b= 6% = 1.06
    c= (null)
    x= (missing value)

    so i put this:

    800(1.8)^x=1000(1.06)^x

    and i don't know where to go from there...

    *not that this is in the functions and relations section, and we are using logs (not sure if that is needed for this question...)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie jessyc's Avatar
    Joined
    Jun 2007
    Posts
    18
    i'v never used  A = P \left(1 + \frac{r}{n}\right)^{nt} in my math class before, so to me it doesn't make much sense...

    (also i tried it out, and the answer doesn't work out for me...)

    i don't think that will work for what i'm trying to do.
    since when i plug
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by jessyc View Post
    i have a question here and i'm not sure how to tackle this sucker...

    Same deposits $800 into an investment fund that erns 8% per year, compounded annually. Jessica deposits $1000 into an investment fund that earns 6% per year, compounded annually. When will their investments be equal( 3 places after the decimal)?

    thats the question and heres what i wrote down...

    Sam
    a= 800
    b= 8% = 1.08
    c= (null)
    x= (missing value)

    Jessica
    a= 1000
    b= 6% = 1.06
    c= (null)
    x= (missing value)

    so i put this:

    800(1.8)^x=1000(1.06)^x

    and i don't know where to go from there...

    *not that this is in the functions and relations section, and we are using logs (not sure if that is needed for this question...)
    First: you have 1.8 where you should have 1.08.

    Second: Take logs of:

    <br />
800(1.08)^x=1000(1.06)^x<br />

    to get:

    <br />
\log(800) + x \log(1.08) = \log(1000) + x \log(1.06)<br />

    so:

    <br />
x (\log(1.08)-\log(1.06)) = \log(1000) - \log(800)<br />

    simplifying:

    <br />
x \log(1.08/1.06) = \log(5/4)<br />

    so:

    <br />
x = \log(5/4)/\log(1.08/1.06) \approx 11.9 \mbox{ years}<br />

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie jessyc's Avatar
    Joined
    Jun 2007
    Posts
    18

    Thumbs up

    Quote Originally Posted by CaptainBlack View Post
    First: you have 1.8 where you should have 1.08.

    Second: Take logs of:

    <br />
800(1.08)^x=1000(1.06)^x<br />

    to get:

    <br />
\log(800) + x \log(1.08) = \log(1000) + x \log(1.06)<br />

    so:

    <br />
x (\log(1.08)-\log(1.06)) = \log(1000) - \log(800)<br />

    simplifying:

    <br />
x \log(1.08/1.06) = \log(5/4)<br />

    so:

    <br />
x = \log(5/4)/\log(1.08/1.06) \approx 11.9 \mbox{ years}<br />

    RonL
    yes sorry about my typo of 1.8, i have written down 1.08. i just had no idea what to do to get x alone. Thanks very much! you have saved my assignment and maybe me passing the course..
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by jessyc View Post
    i'v never used  A = P \left(1 + \frac{r}{n}\right)^{nt} in my math class before, so to me it doesn't make much sense...

    (also i tried it out, and the answer doesn't work out for me...)

    i don't think that will work for what i'm trying to do.
    since when i plug
    you actually ended up using the same formula without trealizing it. now you are correct so far. now all that remains is to find t. we need logs for this. and you should remember, whenever you see a variable as a power and you need to solve for the variable, chances are you need logs.

    let's pick up where you left off.

    800(1.08)^t = 1000(1.06)^t .....divide both sides by 1000

    \Rightarrow \frac {4}{5} (1.08)^t = 1.06^t ......take log base 10 of both sides

    \Rightarrow \log \left( \frac {4}{5} (1.08)^t \right) = \log 1.06^t

    \Rightarrow \log \left( \frac {4}{5} \right) + \log (1.08)^t = \log 1.06^t ...since \log_a xy = \log_a x + \log_a y

    \Rightarrow \log (1.08)^t - \log (1.06)^t = - \log \frac {4}{5}......group t's on one side

    \Rightarrow \log \left( \frac {1.08}{1.06} \right)^t = - \log \frac {4}{5} ......since \log_a \left( \frac {x}{y} \right) = \log_a x - \log_a y

    \Rightarrow t \log \left( \frac {1.08}{1.06} \right) = - \log \frac {4}{5} ......since \log_a x^n = n \log_a x

    \Rightarrow t = - \frac { \log \frac {4}{5}}{ \log \left(  \frac {1.08}{1.06}\right)}

    And now just plug that into your calculator

    EDIT: Ah, CaptainBlack beat me to it, oh well
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member tukeywilliams's Avatar
    Joined
    Mar 2007
    Posts
    307
    I did it slightly differently. Mine might be a simpler method.
    So you use the formula  A = P \left(1 + \frac{r}{n}\right)^{nt} .

    So  800(1.08)^{t} = 1000(1.06)^{t}

     (1.08)^{t} = \frac{5}{4}(1.06)^{t} .

     \frac{(1.08)^{t}}{(1.06)^{t}} = \frac{5}{4}

     \left(\frac{1.08}{1.016} \right)^{t} = \frac{5}{4}

     (1.0188)^{t} = \frac{5}{4}

     \log_{1.0188}  (1.0188)^{t} = \log_{1.0188} 1.25

     t = \log_{1.0188} 1.25 = \frac{\log_{10} 1.25}{\log_{10} 1.0188} = 11.937

    you get  t = 11.937 years.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie jessyc's Avatar
    Joined
    Jun 2007
    Posts
    18
    Quote Originally Posted by Jhevon View Post
    you actually ended up using the same formula without trealizing it. now you are correct so far. now all that remains is to find t. we need logs for this. and you should remember, whenever you see a variable as a power and you need to solve for the variable, chances are you need logs.

    let's pick up where you left off.

    800(1.08)^t = 1000(1.06)^t .....divide both sides by 1000

    \Rightarrow \frac {4}{5} (1.08)^t = 1.06^t ......take log base 10 of both sides

    \Rightarrow \log \left( \frac {4}{5} (1.08)^t \right) = \log 1.06^t

    \Rightarrow \log \left( \frac {4}{5} \right) + \log (1.08)^t = \log 1.06^t ...since \log_a xy = \log_a x + \log_a y

    \Rightarrow \log (1.08)^t - \log (1.06)^t = - \log \frac {4}{5}......group t's on one side

    \Rightarrow \log \left( \frac {1.08}{1.06} \right)^t = - \log \frac {4}{5} ......since \log_a \left( \frac {x}{y} \right) = \log_a x - \log_a y

    \Rightarrow t \log \left( \frac {1.08}{1.06} \right) = - \log \frac {4}{5} ......since \log_a x^n = n \log_a x

    \Rightarrow t = - \frac { \log \frac {4}{5}}{ \log \left(  \frac {1.08}{1.06}\right)}

    And now just plug that into your calculator

    EDIT: Ah, CaptainBlack beat me to it, oh well
    don't fret! yours is still helpfull
    (your a good teacher )

    @ tukeywilliams: if later on i need a simple way of doing it, i'll have to come back and see this
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum