1. ## Logarithm

hey, i have a question and also the answers,
ive worked it out and have the first answer but i just get my head round to how the 2nd answer is....

this is the q..
given that y=ax^n + 3, that y=4.4 when x=10 and y=12.8 when x=100, find the value of 'n' and 'a'

the 2 answers are 0.845 & 0.2

this is my working out for the first answer....

y = ax^n +3 => y=4.4, x=10

4.4 = a10^n + 3

a = (4.4 - 3) / 10^n

= 1.4/10^n

y = 12.8, x = 100

12.8 = a100^n + 3

12.8 = (1.4/10^n) x 100^n + 3

so 100^n = 10^2n

12.8 = (1.4/10^n) x 10^2n + 3

9.8 = 1.4 x 10^n

10^n = 7

log7 = log10^n

n = (log7)/(log10)

= 0.845

i am so sorry if this is long, im currently in 6th form, studying IB
& in order for me to learn, i like having the steps in my working out rite...

so how would i get to the answer 0.2?

2. Originally Posted by exclusiive
hey, i have a question and also the answers,
ive worked it out and have the first answer but i just get my head round to how the 2nd answer is....

this is the q..
given that y=ax^n + 3, that y=4.4 when x=10 and y=12.8 when x=100, find the value of 'n' and 'a'

the 2 answers are 0.845 & 0.2

this is my working out for the first answer....

y = ax^n +3 => y=4.4, x=10

4.4 = a10^n + 3

a = (4.4 - 3) / 10^n

= 1.4/10^n

y = 12.8, x = 100

12.8 = a100^n + 3

12.8 = (1.4/10^n) x 100^n + 3

so 100^n = 10^2n

12.8 = (1.4/10^n) x 10^2n + 3

9.8 = 1.4 x 10^n

10^n = 7

log7 = log10^n

n = (log7)/(log10)

= 0.845

i am so sorry if this is long, im currently in 6th form, studying IB
& in order for me to learn, i like having the steps in my working out rite...

so how would i get to the answer 0.2?
You have the value of $n$, so just substitute it in any equation which contains $n$ and $a$