How do i solve this type of question in Maths Pure year 1?

The questions are from Dr Frost, for example here is a questions:

Solve: 2^(2x-9)= 3^(-7x+1)

Giving your answer in the form: x = Ln a/ Ln b

the answer to this is x= Ln 1536/Ln8748

I'm so confused on how to do this

Reply 1

tonyiptony

11

It would be helpful if we know where exactly you got stuck.
There is a very obvious first step here, but just in case you don't know (all good btw)...

Spoiler

(edited 6 months ago)

Reply 2

DukeAim

OP

7

Original post by tonyiptony

It would be helpful if we know where exactly you got stuck.
There is a very obvious first step here, but just in case you don't know (all good btw)...

Spoiler

I understand how to turn the (2x-9) and (-7x+1) into multiples of Ln (so like putting it behind the Ln) but next in the mark scheme it says "collect the x terms" and that's where i get completely lost

Reply 3

tonyiptony

11

Original post by DukeAim

I understand how to turn the (2x-9) and (-7x+1) into multiples of Ln (so like putting it behind the Ln) but next in the mark scheme it says "collect the x terms" and that's where i get completely lost

Okay, so you got (2x-9)*ln(2)= (-7x+1)*ln(3).
Recall ln(2) and ln(3) is just a number. Expanding brackets and collecting like terms still work, albeit a bit messy.

It's not going to be nice unfortunately.

(edited 6 months ago)

Reply 4

mqb2766

21

Original post by DukeAim

The questions are from Dr Frost, for example here is a questions:

Solve: 2^(2x-9)= 3^(-7x+1)

Giving your answer in the form: x = Ln a/ Ln b

the answer to this is x= Ln 1536/Ln8748

I'm so confused on how to do this

Equivalent to taking logs you can work direct on the indices if youre unsure so
2^2x * 2^-9 = 3^-7x * 3^1
so multilpy throgh by 3^7x and 2^9 (equivlaent to collecting like terms
(2^2 * 3^7)^x = 3 * 2^9
Then evaulate and take logs. Possibly simpler but the main thing is it helps to think about the log steps.

(edited 6 months ago)

How do i solve this type of question in Maths Pure year 1?

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