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Logarithms homework question.

Hi, a question has me trumped on my homework, it says to evaluate:
2log⁡282^{\log_2 8}

My working:
2log⁡282^{\log_2 8}
=22x=8=2^{2^x=8}
=2log2x=log8=2^{log 2^x=log 8}
=2xlog2=log8=2^{xlog2=log8}
=2x=log8log2=2^{x=\frac{log8}{log2}}
=23=2^{3}
=8= 8

I believe I did work out the correct answer of 8, but I'm not sure solving the logarithm as an index of 2 is the correct way to go about it. Has anybody any advice on the proper way to go about solving this? Thanks. :smile:
The main problem is, of course, to evaluate log28 - once you've done that it's a simple matter to raise two to that power.

To compute log28, you want to find x such that 2x=8, and this appears to be what you've done even if your notation is somewhat weird. You should define what x is before introducing it to an equation, and you shouldn't "nest" equations like you've done.

A clearer and more correct way to write what you've done is the following:

let x=log28x=log_2 8
then 2x=8 2^x=8
and log(2x)=xlog(2)=log(8) log(2^x)=xlog(2)=log(8)
which gives x=log(8)log(2)=3x=\frac{log(8)}{log(2)}=3 (alternatively you could simply spot that x=3 from line 2)
therefore 2log28=23=82^{log_2 8}=2^3=8


Overall, however, the simplest thing to do is to notice that log2 is the inverse function of raising 2 to a power, and hence
2log2a=a2^{log_2 a}=a for any a.

In fact, in general,

blogba=ab^{log_b a}=a for any a and any b
(edited 10 years ago)
Reply 2
Avatar for Marcum
Marcum
OP
11
Could anybody help me with this question please? Any pointers?
Given that:
loga(10x−4)=loga(x)+3loga2log_a(10x-4)=log_a(x)+3log_a2
find the value of x.

I know I'm supposed to show that 3loga23log_a2 becomes loga8log_a8, but I don't know where to go from there, any pointers? Thanks.
Reply 3
Avatar for james22
james22
16
Original post by Marcum
Could anybody help me with this question please? Any pointers?
Given that:
loga(10x−4)=loga(x)+3loga2log_a(10x-4)=log_a(x)+3log_a2
find the value of x.

I know I'm supposed to show that 3loga23log_a2 becomes loga8log_a8, but I don't know where to go from there, any pointers? Thanks.


log(x)+log(y)=log(xy)
(edited 10 years ago)
Original post by james22
log(x)+log(y)=log(x)+log(y).


Don't you mean log(x)+log(y)=log(xy)?
Reply 5
Avatar for james22
james22
16
Original post by brianeverit
Don't you mean log(x)+log(y)=log(xy)?


Lol yes it is, although what I wrote was technically true :tongue:
You can also use
Log(x) - log(y) = log(x/y)


Posted from TSR Mobile
Which is incidentally an equivalent rule!

Spoiler

(edited 10 years ago)
Reply 8
Avatar for Marcum
Marcum
OP
11
Sorry I was eating my tea. I got x=2x=2. Is that correct? :smile:
Reply 9
Avatar for james22
james22
16
Original post by Marcum
Sorry I was eating my tea. I got x=2x=2. Is that correct? :smile:


yes
Yes :smile:

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