Damian invested $880 in an account paying an interest rate of 5.2% compounded annually. Assuming no deposits or withdrawals are made, how long would it take, to the nearest tenth of a year, for the value of the account to reach $1,290?
Answers
Answer:
Step-by-step explanation:
A=P(1+r)^t
A=P(1+r)
t
A=1290\hspace{35px}P=880\hspace{35px}r=0.052
A=1290P=880r=0.052
Given values
1290=
1290=
\,\,880(1+0.052)^{t}
880(1+0.052)
t
Plug in values
1290=
1290=
\,\,880(1.052)^{t}
880(1.052)
t
Add
\frac{1290}{880}=
880
1290
=
\,\,\frac{880(1.052)^{t}}{880}
880
880(1.052)
t
Divide by 880
1.4659091=
1.4659091=
\,\,1.052^t
1.052
t
\log\left(1.4659091\right)=
log(1.4659091)=
\,\,\log\left(1.052^t\right)
log(1.052
t
)
Take the log of both sides
\log\left(1.4659091\right)=
log(1.4659091)=
\,\,t\log\left(1.052\right)
tlog(1.052)
Bring exponent to the front
\frac{\log\left(1.4659091\right)}{\log\left(1.052\right)}=
log(1.052)
log(1.4659091)
=
\,\,\frac{t\log\left(1.052\right)}{\log\left(1.052\right)}
log(1.052)
tlog(1.052)
Divide both sides by log(1.052)
7.5449219=
7.5449219=
\,\,t
t
Use calculator
t\approx
t≈
\,\,7.5
7.5
Answer:
Step-by-step explanation:
in deltamath is 7.5