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Algebra 22 Online
OpenStudy (anonymous):

HELP ME DO THE NATURAL LOG PLEASE! If Tanisha has $1000 to invest at 6% per annum compounded monthly, how long will it be before she has $1800? If the compound is continuous, how long will it be? compounding monthly, it will be about __ years before Tanisha has $1800 Compounding continuously, it will be about ___ years before Tanisha has $1800

OpenStudy (anonymous):

\[\frac{ 1800 }{ 1000 }= \left( 1+\frac{ 0.06 }{ 365 } \right)^{365 t} \] That is how far as I got. can someone help me do the natural log I alwasy get stuck there

OpenStudy (phi):

Your equation looks like compounded *daily* can you change it to compounded monthly ?

OpenStudy (anonymous):

opps \[\frac{ 1800 }{ 1000 }= \left( 1+\frac{ 0.06 }{ 12 } \right)^{12 t}\]

OpenStudy (phi):

next, can you simplify some of those numbers? what is 1+0.06/12 ? what is 1800/1000 ?

OpenStudy (anonymous):

\[\frac{ 9 }{ 5 }=\left( 1.005 \right)^{12 t}\]

OpenStudy (phi):

ok, but I would use decimals \[ (1.005)^{12t}= 1.8 \] now take the log (base 10) or ln (log base e) of both sides (either will work, as long as you use the same base for both sides). Example \[ \ln\left(1.005^{12t}\right) = \ln(1.8)\] now use this rule: \[ \ln(a^b) = b \ln(a) \] to rewrite the equation as \[ 12t \ \ln\left(1.005\right) = \ln(1.8)\]

OpenStudy (phi):

now solve for t by dividing both sides by 12 ln(1.005) you will need a calculator to find t (in years)

OpenStudy (anonymous):

would it be 30.07? if I am doing it right.

OpenStudy (phi):

I don't get 30.07 what do you get for ln(1.8) ? then divide by ln(1.005) then divide by 12

OpenStudy (anonymous):

I didnt do ln(1.8) oops . but would you get 9.82. for the continuously it would be the same process. \[1800=1000 e ^{0.06t} \] is how the equation starts.

OpenStudy (phi):

Yes, 9.821 is what I get for compounding monthly for compounding continuously, it is the same process divide both sides by 1000: \[ e^{0.06t}= 1.8 \] take the ln of both sides \[ \ln\left(e^{0.06t}\right)= \ln(1.8) \] use the same "rule" to get \[ 0.06t\ \ln(e)= \ln(1.8) \] now solve for t (you should not need a calculator to figure out ln(e) is 1)

OpenStudy (anonymous):

i got 9.796

OpenStudy (phi):

yes, very close to compounding monthly. a difference of about 0.025 years, or 0.3 months or about 9 days faster for continuous compounding

OpenStudy (anonymous):

Thank you!!! soo much for helping me with the steps!

OpenStudy (phi):

yw

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