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C2 Maths Help!

Solve each of the following equations:

[number] = base

log[3]X + log[3]5 = log[3](2X+3)

log[4]X - log[4](X-1) = log[4]3 + 1/2

2log[6]X = log[6](2x-5) + log[6]5

I would appreciate if you can help me with these questions.
I can do then part way, but then I am not sure of how to find 'X'.
Can you cancel log's?

Thanks.
Rickstahhh

Reply 1

games211

Original post by Rickstahhh

for the first one you use the multiplication rule..and then get rid of the logs

Reply 2

Rickstahhh

Original post by games211

for the first one you use the multiplication rule..and then get rid of the logs

Can you just cancel both logs on either side of the '=' sign?
As I am unsure as we don't seem to have covered that in college.

Reply 3

games211

Original post by Rickstahhh

Can you just cancel both logs on either side of the '=' sign?
As I am unsure as we don't seem to have covered that in college.

after multiplying cant you divide them.. by bringing the right hand-side over..

Reply 4

username893430

Original post by Rickstahhh

First one: use addition rule to get log[3]5x, cancel logs (yep you can do that), rearrange linear equation

Second (trickier): subtraction rule to get log[4]x/x-1, subtract log[4]3 from right hand side, subtraction rule again, then rearrange into a^x=b form

Third: move log[6](2x-5) to left hand side, subtraction rule, cancel logs
edit: didn't notice 2 infront of log[6]X, so it's log[6](x^2)

This is just from me looking at them, so yeah feel free to comment anyone

(edited 10 years ago)

Reply 5

Rickstahhh

Original post by games211

after multiplying cant you divide them.. by bringing the right hand-side over..

For the first question it would like this:

log[3](5X/2X+3)

How would you solve that to work out x?

Reply 6

games211

Original post by Rickstahhh

For the first question it would like this:

log[3](5X/2X+3)

How would you solve that to work out x?

Reply 7

username893430

Original post by Rickstahhh

For the first question it would like this:

log[3](5X/2X+3)

How would you solve that to work out x?

that's another way to go about it (though not one i'd do personally). So you've got

log[3](5X/2X+3)=0

3^0 = 5X/2X+3 (form a^x=b comes from log[a]b=x )

1=5x/2x+3 then a simple rearrangement

Reply 8

Rickstahhh

Original post by adam132

that's another way to go about it (though not one i'd do personally). So you've got

log[3](5X/2X+3)=0

3^0 = 5X/2X+3 (form a^x=b comes from log[a]b=x )

1=5x/2x+3 then a simple rearrangement

Original post by games211

Thanks a lot, it makes sense now

Reply 9

Rickstahhh

How would you solve this equation:
2^2y+1 + 7(2^y) - 15 = 0

So you would let y=2^y

Which gives y^2 + 7y - 15 = 0

But cannot be factorised.
I think I have done something wrong in the previous steps

Reply 10

username893430

Original post by Rickstahhh

How would you solve this equation:
2^2y+1 + 7(2^y) - 15 = 0

So you would let y=2^y

Which gives y^2 + 7y - 15 = 0

But cannot be factorised.
I think I have done something wrong in the previous steps

Ahh well 2^(2y+1) is actually (2^1)2^2y
Since (a^x)*(a^1) = a^(a+1)
(general index laws, apologies if looks confusing)
So it's actually 2y^2

Posted from TSR Mobile

Reply 11

Rickstahhh

Original post by adam132

Ahh well 2^(2y+1) is actually (2^1)2^2y
Since (a^x)*(a^1) = a^(a+1)
(general index laws, apologies if looks confusing)
So it's actually 2y^2

Posted from TSR Mobile

Thanks, I actually figured it out eventually