Clueless as what to do here...

Two variables, x and y, are connected by the law . The graph of log to the base 4 y against x is a straight line passing through the origin and the point A(6,3). Find the value of a.

## 2006 paper 1 question 10

### BuckminsterFullerene

Posted 14 May 2007 - 09:05 PM

Hello Amazoph

Firstly

For all questions as such, you begin with what they've given you ie y=a^x. They have also told you that their graph is of log (base) 4 and x.

Hence, you should, take the log(base) 4 of both sides

Ie. Log(4)y=Log(4)a^x

Log (4)y = x . log(4)a

You should now recognise that log(4)a is the gradient because this is similar to y=mx+c

hence;

You know the gradient of the line is 0.5

so

Log(4)a=0.5

Hence

4^0.5 = a

a = 2

Glad to help!

Firstly

For all questions as such, you begin with what they've given you ie y=a^x. They have also told you that their graph is of log (base) 4 and x.

Hence, you should, take the log(base) 4 of both sides

Ie. Log(4)y=Log(4)a^x

Log (4)y = x . log(4)a

You should now recognise that log(4)a is the gradient because this is similar to y=mx+c

hence;

You know the gradient of the line is 0.5

so

Log(4)a=0.5

Hence

4^0.5 = a

a = 2

Glad to help!

### antollo

Posted 17 April 2008 - 09:51 PM

Hello Amazoph

Firstly

For all questions as such, you begin with what they've given you ie y=a^x. They have also told you that their graph is of log (base) 4 and x.

Hence, you should, take the log(base) 4 of both sides

Ie. Log(4)y=Log(4)a^x

Log (4)y = x . log(4)a

You should now recognise that log(4)a is the gradient because this is similar to y=mx+c

hence;

You know the gradient of the line is 0.5

so

Log(4)a=0.5

Hence

4^0.5 = a

a = 2

Glad to help!

Firstly

For all questions as such, you begin with what they've given you ie y=a^x. They have also told you that their graph is of log (base) 4 and x.

Hence, you should, take the log(base) 4 of both sides

Ie. Log(4)y=Log(4)a^x

Log (4)y = x . log(4)a

You should now recognise that log(4)a is the gradient because this is similar to y=mx+c

hence;

You know the gradient of the line is 0.5

so

Log(4)a=0.5

Hence

4^0.5 = a

a = 2

Glad to help!

hi,

i follow how u get the gradient to equal log(4)a,

but im confused how you got the gradient of the line to be 0.5 =/

it's probably something really simple, but i cant seem to figure it out