y=ax+b law of indices question

Are you sure you got the question right for the first one?

For the second, use the rules for indices.

EDIT: For the first one I'm guessing there should be some brackets in there. Do you mean

6^{y+1}=36^{x-2}

?

Reply 3

OP

Original post by notnek

You need to use brackets but I'm guessing you mean

6^{y+1}=36^{x-2}

?

You need to use the fact that

36=6^2

so

6^{y+1}=(6^2)^{x-2}

Next use the index law

(a^b)^c = a^{b\times c}

to change the right-hand-side.

Post all your working if you get stuck.

Yes I forgot to put in the brackets

6^(y+1)=6^2(x-2) is not quite in the form of y=ax+b, so shall I move the numbers to one side or expand (6^2)^x-2 to = 6^2x-4?

For b) because-1/2y is negative I would need to change that positive before I apply indices law?

Reply 4

cambo211

19

Original post by adli93

Yes I forgot to put in the brackets

6^(y+1)=6^2(x-2) is not quite in the form of y=ax+b, so shall I move the numbers to one side or expand (6^2)^x-2 to = 6^2x-4?

For b) because-1/2y is negative I would need to change that positive before I apply indices law?

Bolded.

Reply 5

Notnek

21

Original post by adli93

Yes I forgot to put in the brackets

6^(y+1)=6^2(x-2) is not quite in the form of y=ax+b, so shall I move the numbers to one side or expand (6^2)^x-2 to = 6^2x-4?

For b) because-1/2y is negative I would need to change that positive before I apply indices law?

You need to expand as you say so you have

6^{y+1}=6^{2x-4}

Since both sides are powers of 6, can you see what you can do next?

You still need to use more brackets when posting. See here for more info.

Reply 6

OP

Original post by notnek

You need to expand as you say so you have

6^{y+1}=6^{2x-4}

Since both sides are powers of 6, can you see what you can do next?

You still need to use more brackets when posting. See here for more info.

Not really sure.. group the 6's to one side. But what would I do with ^(y+1) and ^(2x-4). How can I type the equation like you can? Would help you and others view it easier

Reply 7

Original post by adli93

Not really sure.. group the 6's to one side. But what would I do with ^(y+1) and ^(2x-4). How can I type the equation like you can? Would help you and others view it easier

If

10^a=10^b

, what can you say about a and b?

To type equations like that use latex.

Reply 8

OP

Original post by james22

If

10^a=10^b

, what can you say about a and b?

To type equations like that use latex.

Well a and b are powers of 10. What I do not understand is what to do because the = sign is there.

Reply 9

Original post by adli93

Well a and b are powers of 10. What I do not understand is what to do because the = sign is there.

a and b need not be powers of 10, you could have a=2 for example.

Would it be possible to have a=2, b=3? What about a=5, b=7? What link is there between a and b?

Reply 10

OP

Original post by james22

a and b need not be powers of 10, you could have a=2 for example.

Would it be possible to have a=2, b=3? What about a=5, b=7? What link is there between a and b?

My mistake. Yes it would be possible. I would say they are numbers. Referring to the question, a and b are (y+1), (x-2) respectively

Reply 11