How do you solve $120 = 100 {(1 + (\frac{.032}{12}))}^{12 t}$ $120=100(1+(.032/12))^(12t)$ ?

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How do you solve $120 = 100 {(1 + (\frac{.032}{12}))}^{12 t}$ $120=100(1+(.032/12))^(12t)$ ?

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Answer 1 · 2015-11-14T03:49:11

Use logarithms.

Explanation:

Divide both sides by 100: $1.2 = {(1 + (\frac{.032}{12}))}^{12 t}$ $1.2=(1+(.032/12))^(12t)$
Simplify the inside: $1.2 = {(1.002 ¯ 6)}^{12 t}$ $1.2=(1.002bar6)^(12t)$
Convert into a logarithm: ${log}_{1.2} 1.002 ¯ 6 = 12 t$ $log_"1.2"1.002bar6=12t$
Change of base formula: $\frac{log 1.002 ¯ 6}{log 1.2} = 12 t$ $(log1.002bar6)/log1.2=12t$
Divide both sides by 12: $\frac{log 1.002 ¯ 6}{12 log 1.2} = t$ $color(blue)((log1.002bar6)/(12log1.2)=t$

Answer 2 · 2015-11-14T03:52:53

$t = \frac{ln (\frac{6}{5})}{12 ln (1 + \frac{.032}{12})}$ $t= ln(6/5)/(12ln(1+.032/12))$

Explanation:

We will be using the property of logarithms that
$ln (x^{n}) = n ln (x)$ $ln(x^n) = nln(x)$
(this is very useful for solving for variables in exponents)

$120 = 100 {(1 + \frac{.032}{12})}^{12 t}$ $120 = 100(1+.032/12)^(12t)$

$\Rightarrow \frac{120}{100} = \frac{6}{5} = {(1 + \frac{.032}{12})}^{12 t}$ $=>120/100= 6/5 = (1+.032/12)^(12t)$

$\Rightarrow ln (\frac{6}{5}) = ln ({(1 + \frac{.032}{12})}^{12 t})$ $=>ln(6/5) = ln((1+.032/12)^(12t))$

$\Rightarrow ln (\frac{6}{5}) = 12 t ln (1 + \frac{.032}{12})$ $=> ln(6/5) = 12tln(1+.032/12)$ (by the property stated above)

Now, solving for $t$ $t$ gives

$t = \frac{ln (\frac{6}{5})}{12 ln (1 + \frac{.032}{12})}$ $t = ln(6/5)/(12ln(1+.032/12))$

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How do you solve $120 = 100 {(1 + (\frac{.032}{12}))}^{12 t}$ $120=100(1+(.032/12))^(12t)$ ?

2 Answers

Explanation:

Explanation:

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How do you solve 120=100(1+(.03212))12t120=100(1+(.032/12))^(12t)?

2 Answers

Explanation:

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How do you solve $120 = 100 {(1 + (\frac{.032}{12}))}^{12 t}$ $120=100(1+(.032/12))^(12t)$ ?