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Kiiten superhelp thread

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Reply 40
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kiiten
OP
18
Log question - express in terms of logax , logay, logaz

Ive attached the ques h.) & my working but i dont understand where the 5/2 comes from.

Answer is 5/2 + 1/2logax - 3logay + 5/2logaz

Posted from TSR Mobile
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Reply 41
Avatar for Zacken
Zacken
22
di
Original post by kiiten
Log question - express in terms of logax , logay, logaz

Ive attached the ques h.) & my working but i dont understand where the 5/2 comes from.

Answer is 5/2 + 1/2logax - 3logay + 5/2logaz

Posted from TSR Mobile


log⁡24=log⁡222=2log⁡22=2×1=2\log_2 4 = \log_2 2^2 = 2\log_2 2 = 2 \times 1 = 2

log⁡aaxz5y3=log⁡aaxz5−log⁡ay3\displaystyle \log_a \frac{\sqrt{axz^5}}{y^3} = \log_a \sqrt{axz^5} - \log_a y^3

Now log⁡ay3=3log⁡ay\log_a y^3 = 3\log_a y

and log⁡aaxz5=12log⁡a(axz5)=12(log⁡aa+log⁡ax+log⁡az5)\log_a \sqrt{axz^5} = \frac{1}{2} \log_a (axz^5) = \frac{1}{2}\left(\log_a a + \log_a x + \log_a z^5\right)

Can you take it from here? What's log⁡aa\log_a a?
Reply 42
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kiiten
OP
18
Original post by Zacken
di

log⁡24=log⁡222=2log⁡22=2×1=2\log_2 4 = \log_2 2^2 = 2\log_2 2 = 2 \times 1 = 2

log⁡aaxz5y3=log⁡aaxz5−log⁡ay3\displaystyle \log_a \frac{\sqrt{axz^5}}{y^3} = \log_a \sqrt{axz^5} - \log_a y^3

Now log⁡ay3=3log⁡ay\log_a y^3 = 3\log_a y

and log⁡aaxz5=12log⁡a(axz5)=12(log⁡aa+log⁡ax+log⁡az5)\log_a \sqrt{axz^5} = \frac{1}{2} \log_a (axz^5) = \frac{1}{2}\left(\log_a a + \log_a x + \log_a z^5\right)

Can you take it from here? What's log⁡aa\log_a a?

How did you get to -logay^3 on the second line of working ?
Reply 43
Avatar for Zacken
Zacken
22
Original post by kiiten
How did you get to -logay^3 ?


log⁡ab=log⁡a−log⁡b\log \frac{a}{b} = \log a - \log b except in this case, we have a=axz5a =\sqrt{axz^5} and b=y3b = y^3
Reply 44
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kiiten
OP
18
Original post by Zacken
log⁡ab=log⁡a−log⁡b\log \frac{a}{b} = \log a - \log b except in this case, we have a=axz5a =\sqrt{axz^5} and b=y3b = y^3


Umm i think ive done it wrong but i got
2 + 1/2loga + 1/2logx + 5/2logz - 3logay
Reply 45
Avatar for Zacken
Zacken
22
Original post by kiiten
Umm i think ive done it wrong but i got
2 + 1/2loga + 1/2logx + 5/2logz - 3logay


Yeah, that's correct, just remember that 12log⁡aa=12×1=12\frac{1}{2} \log_a a = \frac{1}{2} \times 1 = \frac{1}{2} so you can simplify your first two terms as 2+12=⋯2 + \frac{1}{2} = \cdots
Reply 46
Avatar for kiiten
kiiten
OP
18
Original post by Zacken
Yeah, that's correct, just remember that 12log⁡aa=12×1=12\frac{1}{2} \log_a a = \frac{1}{2} \times 1 = \frac{1}{2} so you can simplify your first two terms as 2+12=⋯2 + \frac{1}{2} = \cdots


So 5/2 + 1/2logax + 5/2logaz - 3logay
Reply 47
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Zacken
22
Original post by kiiten
So 5/2 + 1/2logax + 5/2logaz - 3logay


Seems fine, yups.
Reply 48
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kiiten
OP
18
Thank you :smile:

If you have log (1+5^-x) would it simplify to -xlog6 ?
Reply 49
Avatar for Zacken
Zacken
22
Original post by kiiten
Thank you :smile:

If you have log (1+5^-x) would it simplify to -xlog6 ?


No, it wouldn't; you can't simplify that anymore. Remember that log⁡(a+b)≠log⁡a+log⁡b\log(a+b) \neq \log a+ \log b. :-)
Reply 50
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kiiten
OP
18
Original post by Zacken
No, it wouldn't; you can't simplify that anymore. Remember that log⁡(a+b)≠log⁡a+log⁡b\log(a+b) \neq \log a+ \log b. :-)


So how would you solve
Log (1×5^-x) = log3
Reply 51
Avatar for Zacken
Zacken
22
Original post by kiiten
So how would you solve
Log (1×5^-x) = log3


"cancel" the logs: 1+5−x=3⇒5−x=2⇒5x=121 + 5^{-x} = 3 \Rightarrow 5^{-x} = 2 \Rightarrow 5^x = \frac{1}{2} - now you know how to solve this using logarithms.
Reply 52
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kiiten
OP
18
Original post by Zacken
"cancel" the logs: 1+5−x=3⇒5−x=2⇒5x=121 + 5^{-x} = 3 \Rightarrow 5^{-x} = 2 \Rightarrow 5^x = \frac{1}{2} - now you know how to solve this using logarithms.


Oh i see - could you do that for any log e.g. log5^-x = log3 instead of simplifying
Reply 53
Avatar for Zacken
Zacken
22
Original post by kiiten
Oh i see - could you do that for any log e.g. log5^-x = log3 instead of simplifying


If you have log⁡a=log⁡b\log a = \log b - then you know that a=ba=b.

If you have log⁡a+log⁡b=log⁡c\log a + \log b = \log c IT IS NOT TRUE THAT a+b=ca+b = c BUT IT IS TRUE that log⁡a+log⁡b=log⁡ab=log⁡c⇒ab=c\log a + \log b = \log ab = \log c \Rightarrow ab=c

Basically when you have log equations you should try and get it in the form log⁡(something)=log⁡(something else)\log (\text{something}) = \log (\text{something else}) then you can say something = something else. (the bases of the logs have to be the same, if they're not, you can always use change of base rule) .
Reply 54
Avatar for kiiten
kiiten
OP
18
Original post by Zacken
If you have log⁡a=log⁡b\log a = \log b - then you know that a=ba=b.

If you have log⁡a+log⁡b=log⁡c\log a + \log b = \log c IT IS NOT TRUE THAT a+b=ca+b = c BUT IT IS TRUE that log⁡a+log⁡b=log⁡ab=log⁡c⇒ab=c\log a + \log b = \log ab = \log c \Rightarrow ab=c

Basically when you have log equations you should try and get it in the form log⁡(something)=log⁡(something else)\log (\text{something}) = \log (\text{something else}) then you can say something = something else. (the bases of the logs have to be the same, if they're not, you can always use change of base rule) .


Thanks that helps to explain things :smile:
Reply 55
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Zacken
22
Original post by kiiten
Thanks that helps to explain things :smile:


No problem.
Reply 56
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kiiten
OP
18
What about this question - im not sure where you would start. Ive attached some working
Attachment not found


1458040767458607763651.jpg
14580409406772006253070.jpg511.4KB
image.png

Use Log rules.
Reply 58
Avatar for kiiten
kiiten
OP
18
Original post by Mystery.
image.png

Use Log rules.


Have i done it right so far. So, 5/ (log3x) = 6 / (log3)
Reply 59
Avatar for Notnek
Notnek
21
Original post by kiiten
What about this question - im not sure where you would start. Ive attached some working
Attachment not found


1458040767458607763651.jpg

You have

log⁡3x+log3y=5\log_3 x + log_3 y = 5

log⁡3x×log3y=6\log_3 x \times log_3 y = 6


Let a=log⁡3xa=\log_3 x and b=log⁡3yb=\log_3 y then you have

a+b=5a+b=5

ab=6ab = 6

So you can ignore logs and solve these equations for aa and bb. Then substitute back in for aa and bb.

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