# Thread: a(b)^x = a(b)^x ( getting x value )

1. ## 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...)

2. 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

3. Originally Posted by jessyc
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:

$
800(1.08)^x=1000(1.06)^x
$

to get:

$
\log(800) + x \log(1.08) = \log(1000) + x \log(1.06)
$

so:

$
x (\log(1.08)-\log(1.06)) = \log(1000) - \log(800)
$

simplifying:

$
x \log(1.08/1.06) = \log(5/4)
$

so:

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

RonL

4. Originally Posted by CaptainBlack
First: you have $1.8$ where you should have $1.08$.

Second: Take logs of:

$
800(1.08)^x=1000(1.06)^x
$

to get:

$
\log(800) + x \log(1.08) = \log(1000) + x \log(1.06)
$

so:

$
x (\log(1.08)-\log(1.06)) = \log(1000) - \log(800)
$

simplifying:

$
x \log(1.08/1.06) = \log(5/4)
$

so:

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

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..

5. Originally Posted by jessyc
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

6. 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.

7. Originally Posted by Jhevon
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