percentage increase word problem help

[shelly89]

An investment plan offers 8% grwth each year. £5799 is invested in the plan.
a) how much is the plan worth after 4 years?
b) after how many years is the investmet worth approximately double its intial value.?

my working.....


100% + 8% = 108%

1.08^{4} x 5799= £7889

is this correct?
for the second part am not sure what to d0?

5799 x 2 = 11598

so after how many earyears will it take to ge to this value?

is the equation

1.08^{x} = 11598?

 

[JeffM]

An investment plan offers 8% grwth each year. £5799 is invested in the plan.
a) how much is the plan worth after 4 years?
b) after how many years is the investmet worth approximately double its intial value.?

my working.....


100% + 8% = 108%

1.08^{4} x 5799= £7889

is this correct? Spot on if you are not worried about pence
for the second part am not sure what to d0?

5799 x 2 = 11598 Yes

so after how many earyears will it take to ge to this value?

is the equation

1.08^{x} = 11598? Yes indeed although you should remember that the question asks for an approximate answer.

So what is your problem exactly? Do you not know how to solve your equation? Hint: do you remember logs?

 

[shelly89]

So what is your problem exactly? Do you not know how to solve your equation? Hint: do you remember logs?

Hello, thank you for your reply, its just, if the equation to solve is 1.08^{x} = 11598,

than when i solve this for 'x' I get a strange answer like 121....

I think the answer is 9 years, but I got that answer by trial and error,

1.08^{x} = 11598

xln1.08 = ln11598

x = ln11598/ln1.08?

Is there an easier way to do this question, other than using logs?

 

[srmichael]

Hello, thank you for your reply, its just, if the equation to solve is 1.08^{x} = 11598,

than when i solve this for 'x' I get a strange answer like 121....

I think the answer is 9 years, but I got that answer by trial and error,

1.08^{x} = 11598 ===> Not quite (see below)

xln1.08 = ln11598

x = ln11598/ln1.08?

Is there an easier way to do this question, other than using logs?

\(\displaystyle 1.08^x(5,799)=11,598\)

or

\(\displaystyle 1.08^x=2\)

Now proceed...

 

[shelly89]

\(\displaystyle 1.08^x(5,799)=11,598\)

or

\(\displaystyle 1.08^x=2\)

Now proceed...

oh thank you, I forgot to multiply by 5799..

 

[JeffM]

Hello, thank you for your reply, its just, if the equation to solve is 1.08^{x} = 11598,

than when i solve this for 'x' I get a strange answer like 121....

I think the answer is 9 years, but I got that answer by trial and error,

1.08^{x} = 11598

xln1.08 = ln11598

x = ln11598/ln1.08?

Is there an easier way to do this question, other than using logs?

OK I answered your question incorrectly. Sorry about that.

Let's go through it. When you wanted to get the value in four years you did

\(\displaystyle 1.08^4 * 5799 \approx 7889.48.\)

So when you want to double it, but your unknown is the number of years.

\(\displaystyle 1.08^x * 5799 \approx 2 * 5799 \implies\)

\(\displaystyle 1.08^x \approx 2 \implies\)

\(\displaystyle log(1.08^x) \approx log(2) \implies\)

\(\displaystyle x * log(1.08) \approx log(2) \implies\)

\(\displaystyle x \approx \dfrac{log(2)}{log(1.08)} \approx 9.\)