Ok, ive done that and I got 27 and 8 for both x and y (4 solutions). 27 is right but 8 is wrong - its supposed to be 9. Where have I gone wrong?

u + v = 5 so u = 5 - v
subbed this into uv = 6 to get v^2 - 5v - 6 =0
so v=3 v=2 which I put into u = 5 - v to get u=2 and u=3
log3x = 2 etc. to get 27 and 8 ?

Reply 66

Zacken

Original post by kiiten

Those are your solutions for x - what about the corresponding ones for y?

Reply 67

kiiten

Original post by Zacken

Those are your solutions for x - what about the corresponding ones for y?

Sorry I meant x=8, x=27 and y=27, y=8. But, the answer is supposed to be 9 not 8.

Reply 68

Zacken

Original post by kiiten

Sorry I meant x=8, x=27 and y=27, y=8. But, the answer is supposed to be 9 not 8.

\log_3 x = 2

then

x = 3^2 = 9

not

x=2^3 = 8

. :-)

In general:

\log_a x = y \Rightarrow x = a^y \neq y^a

Reply 69

kiiten

Original post by Zacken

\log_3 x = 2

then

x = 3^2 = 9

not

x=2^3 = 8

. :-)

In general:

\log_a x = y \Rightarrow x = a^y \neq y^a

Ahh I see its another log rule. Thanks :smile:

Reply 70

kiiten

Referring back to the point mentioned earlier when logs can cancel out if you had
log2 (72) = m+nlog2 (3)
I know n=2 from a previous question so would 72 = 3 ^2+m ?

I'm not sure I'm doing this right but the question is asking to show that log2 (72) = m+nlog2 (3)

Reply 71

Zacken

Original post by kiiten

Referring back to the point mentioned earlier when logs can cancel out if you had
log2 (72) = m+nlog2 (3)
I know n=2 from a previous question so would 72 = 3 ^2+m ?

No! You can only cancel logs when you have only one log on each side.

I'm not sure I'm doing this right but the question is asking to show that log2 (72) = m+nlog2 (3)

Post the question, please.

Reply 72

kiiten

Original post by Zacken

No! You can only cancel logs when you have only one log on each side.

Post the question, please.

1.c)

I worked out n=2 from the previous question. So if you rearrange would it be 2+m = log2(72) / log2(3)

(edited 8 years ago)

Reply 73

Zacken

Original post by kiiten

1.c)

I worked out n=2 from the previous question. So if you rearrange would it be 2+m = log2(72) / log2(3)

All they want you to do is write

\log_2 72 = (8 \times 9) = \log_2 8 + \log_2 9

and then you know that

\log_2 9 = n \log_2 3

and that

\log_2 8 = \log_2 2^3 = 3 \log_2 2 = 3

.

So that you can then write

\log_2 72 = m + n\log_2 3

for

m, n

that you've found.

Reply 74

kiiten

Original post by Zacken

All they want you to do is write

\log_2 72 = (8 \times 9) = \log_2 8 + \log_2 9

and then you know that

\log_2 9 = n \log_2 3

and that

\log_2 8 = \log_2 2^3 = 3 \log_2 2 = 3

.

So that you can then write

\log_2 72 = m + n\log_2 3

for

m, n

that you've found.

Hm ok but why do you need the last part 3log2(2)=3

Reply 75

Zacken

Original post by kiiten

Hm ok but why do you need the last part 3log2(2)=3

Because that's your

m

Reply 76

kiiten

Original post by Zacken

Because that's your

m

Oh so its mlog2(3) + nlog2(3)
Thanks :smile:

Reply 77

Zacken

Original post by kiiten

Oh so its mlog2(3) + nlog2(3)
Thanks :smile:

No! It's

\log_2 72 = \log_2 (8\times 9) = \log_2 8 + 2\log_2 3 = 3 + 2\log_2 3

m=3

and

n=2

Reply 78

kiiten

Original post by Zacken

No! It's

\log_2 72 = \log_2 (8\times 9) = \log_2 8 + 2\log_2 3 = 3 + 2\log_2 3

m=3

and

n=2

Oops so does the original question mean that log2(72) = (m+n)log2(3)
= 2log2(3) + 3log2(2)

Reply 79

kiiten

if you have to determine if a point is maximum of minimum do you use the first or second derivative? Also, how do you know if it is max or min - is it +ve is max?