Emily got a new job that guarantees her a 6% raise every year. If she started out making $25,000, how long will it be before she doubles her current salary?

Guest Apr 8, 2015

#1**+10 **

Let's first take a look at what happens in the first years.

Emily starts out with a salary of $25,000

After one year her salary has increased with 6%.

This means her salary is 106% of what it was last year.

Therefore her salary is now

$$(\frac{106}{100})\times\$25,000 = 1.06 \times\$25,000 = \$26,500$$

One year later her salary again increases with 6%

The means her salary is again 106% of what it was last year

Therefore her salary is now

$$(\frac{106}{100})\times\$26,500 = 1.06 \times \$26,500 = \$28,090$$

Now suppose we had the same situation and we wanted to know what her salary was after 2 years.

Instead of first calculating her salary after one year we could also have immediately calculated

$$(\text{salary after two years}) = 1.06 \times (\text{salary after one year}) = 1.06 \times 1.06 \times (\text{starting salary}) = (1.06)^2 \times(\text{starting salary}) = (1.06)^2 \times \$25,000 = \$28,090$$

Similarly, we could show that for 3 years we have that

$$(\text{salary after three years}) = 1.06^3 (\text{starting salary}) = 1.06^3 \times \$25,000$$

and that for x years we have that

$$(\text{salary after x years}) = 1.06^x = (\text{starting salary})$$

Now, if we rewrite the exercise as an equation we need to solve

$$1.06^x \times \$25,000 = 2 \times \$25,000$$

dividing both sides by 25000 gives us

$$1.06^x = 2$$

Now, I'm not sure what your level is so I'll give you two options

**option 1 (the easy option (for whole years)):**

Start with 1 and keep multiplying by 1.06 until the value is bigger or equal to 2.

The number of times you multiplied is the value of x.

To illustrate, this gives

$$\begin{array}{lcl}

1*1.06 = 1.06\\

1*1.06*1.06 = 1.06^2 = 1.1236\\

1*1.06*1.06*1.06 = 1.06^3 = 1.191016\\

(...)\\

1.06^{11} = 1.898298558\\

1.06^{12} = 2.012196472\\

\end{array}$$

Hence after 12 years, her income has doubled.

**Option 2 (the advanced option (for partial years)):**

This option makes use of logarithms. If you've never heard of logarithms, this method is not for you.

We indicate the natural logarithm as $$ln() = {}^elog()$$

We have that

$$\begin{array}{lcl}

1.06^x = 2\\

e^{ln(1.06^x)} = 2\\

e^{x \times ln(1.06)} = 2\\

x \times ln(1.06) = ln(2)\\

x = \frac{ln(2)}{ln(1.06)} \approx 11.8957

\end{array}$$

Hence the answer is approximately 11.8957 years

Reinout

reinout-g
Apr 8, 2015