Thread: help finishing of a question

i have started this , but unable to find a way to complete it, any hints?

Q: in order to test a model of the form y=ab^x, a student plots a graph of log y against x for her values of x and y. The points lie in a straight line, with the gradient 0.65. the straight line meets the log y axis at the point (0,1.05)
a) find the values of a and b correct to 1.d.p
b) hence find the value of y, given by the model, wehn x=3

Workings so far
y=ab^x
logy=logab^x
logy=loga + xlogb
logy=xlogb +loga => which in relation is y=mx+c
i know that loga=1.05 which is where it crosses the y axis
and that xlogb=0.65 which is the gradient
but here is where i get stuck. i'm not sure how to reverse log or something like that to work out a and b

Originally Posted by JorCherryhead

i know that loga=1.05 which is where it crosses the y axis
and that xlogb=0.65 which is the gradient
but here is where i get stuck. i'm not sure how to reverse log or something like that to work out a and b

You have the right idea, but the gradient is not xlogb, but just logb = 0.65, which you can solve for b. ....etc.

how would you find the value for b by using a scientific calc?

Originally Posted by JorCherryhead

how would you find the value for b by using a scientific calc?

I wouldn't, but I'm old, and have slide rules, and even tables as well as a calculator and two computers [Google will do claculations for you. Just type it in the search window.] However, ...assuming that you are using logs base 10, use your knowledge of logarithms:

If log[10]b = 0.65, then b = 10^0.65

My calculator [which I'll use for this purpose] has a y^x button for this purpose:

10
y^x
0.65
=
and I get 4.4668, which you might round.

thankyou again
yeah using the rule does make things alot simpler. as i was looking at the logb=0.65 and thinking what can i do with this.
so with the last part of the question, is it just simply subsituting all that i have worked out to find the value of y?