1. AS Logarithm Exam Q's

Hi

There are a couple of questions on logs that I'm having some difficulty with.
One of them is typed and attached,

Thanks

2. Originally Posted by Apex
Hi

There are a couple of questions on logs that I'm having some difficulty with.
One of them is typed and attached,

Thanks
You're doing ok so far. now just drop the logs and solve for x, since if the logA = logB then A = B

3. Originally Posted by Jhevon
You're doing ok so far. now just drop the logs and solve for x, since if the logA = logB then A = B
I see-Thanks :-)

4. Hi

I've tried what you mentioned:

(4x+6)/3 = x+4

4x+6 = 3x+12
:.
x = 6
y = 30

The textbook gives the following answers though:
x = 3/22
y = 24/11

??

5. Originally Posted by Apex
Hi

I've tried what you mentioned:

(4x+6)/3 = x+4

4x+6 = 3x+12
:.
x = 6
y = 30

The textbook gives the following answers though:
x = 3/22
y = 24/11

??
Hmm, interesting. I'll try to see what method they used. Apparently there is more than one solution to this thing, since if you check by plugging in the answers we got, you would find that they work out as well (by the way, y = 10 if you do it my way, not 30)

what would happen if we started working on the logs equation first?

6. Originally Posted by Jhevon
Hmm, interesting. I'll try to see what method they used. Apparently there is more than one solution to this thing, since if you check by plugging in the answers we got, you would find that they work out as well (by the way, y = 10 if you do it my way, not 30)

what would happen if we started working on the logs equation first?
strange...
I was just thinking about dropping the logs, I don't think we can do that 'easily' in this case. On the RHS of the second equation provided in the question, "x+4" is not in brackets, :. I don't think the log is associated with that so we probably cannot drop the logs like we did...

7. I've managed to work it out and get textbook answers:

8. Originally Posted by Apex
Hi

I've tried what you mentioned:

(4x+6)/3 = x+4...
Hello,

since the LHS consists of onelogarithm the RHS must be one logarithm to the same base. Thus "dropping the logs" becomes:

(4x+6)/3 = x*16 (Remember: Adding logs is the same as multiplying the numerii)
6/3 = 16x - 4/3x = 44/3 x

x = 6/44 = 3/22

9. Originally Posted by earboth
Hello,

since the LHS consists of onelogarithm the RHS must be one logarithm to the same base. Thus "dropping the logs" becomes:

(4x+6)/3 = x*16 (Remember: Adding logs is the same as multiplying the numerii)
6/3 = 16x - 4/3x = 44/3 x

x = 6/44 = 3/22
Hi
Thanks, I don't understand how you got the 16. I know that
logX + logY = log(XY)...
Can you please expand on that method? (might help in solving other q's)
Thanks

10. First question more or less done, here is another:

11. Originally Posted by Apex
First question more or less done, here is another:
Hello,

one additional remark to the first problem:

16 = 2^4 . So if you know that the sum x+4 is a logarithm to the base 2 you'll get from
2^(x+4) = 2^x * 2^4 = 2^x * 16

to the last problem:
(4th row of your transformations. You have forgotten to use brackets)

(log_{3}x)^2 - log_{3}x = log_{3}9 Use substitution: log_{3}x = y. Then you have:

y^2 - y - 2 = 0 (log_{3}9 = 2) Solve for y. You'll get y = 2 or y = -1

Now re-substitute:
log_{3}x = 2 ==> x = 9

log_{3}x = -1 ==> x = 3^(-1) = 1/3

12. to the last problem:
(4th row of your transformations. You have forgotten to use brackets)

(log_{3}x)^2 - log_{3}x = log_{3}9 Use substitution: log_{3}x = y. Then you have:

y^2 - y - 2 = 0 (log_{3}9 = 2) Solve for y. You'll get y = 2 or y = -1

Now re-substitute:
log_{3}x = 2 ==> x = 9

log_{3}x = -1 ==> x = 3^(-1) = 1/3
Thanks. looks like I have to keep my eyes open for these substitutions, otherwise I'd just solve like I did and get only one ans.

13. Another question,

14. Originally Posted by Apex
Another question,
Change the log_9 to log_3 rather than the other way. Then simplify so
that there is one log_3 on each side then equate what is inside the logs
on both sides to give a quadratic.

RonL

15. Thanks for that ;)