Question

log(base of 2)15 +log(base of 7)x +log(base of 7)y help please.

Answer 1

^_^
Sorry... uhh what are you supposed to do with these?

Answer 2

\[\Large \log_2(15) + \log_7(x) + \log_7(y)\]

Answer 3

^ thats what it is
lol sorry i'm new to this site.

Answer 4

That's what it is, yes, but what are your instructions? LOL

and yes, welcome to Openstudy ^_^

Answer 5

have to write it as a single log

Answer 6

Tricky... one of them is log base 2 LOL

Answer 7

i know man my teacher is insanely hard....Like the hardest out of my whole school......

Answer 8

Relax.. there is 'change of base' formula, have you heard of it?

Answer 9

nope not at all she doesn't like to give us "details" she gives us an equation and tells us to solve.

Answer 10

That must be frustrating, huh :3

Let's let
\[\Large a = \log_2(15)\]
Somehow, we want to turn this into an expression involving only log base 7.

Answer 11

ok, so how would i do that? i have an TI-84 Plus. and know how to type it in
but i dont know how i would exactly change it into base 7

Answer 12

oh wait, would it be 12.676?

Answer 13

Who needs calculators anyway :P
Let's get rid of that log for now, by raising 2 to both sides of the equation:
\[\LARGE \color{red}2^a= \color{red}2^{\log_2(15)}\]

Answer 14

Can you simplify the right-hand-side?

Answer 15

is it 9.031?

Answer 16

Turn off that [bloody] calculator (^_^) and just look at the expression:
\[\LARGE 2^{\log_2(15)}\]
doesn't this remind you of a certain property of logarithms?

Answer 17

haha ok the calculator is off. ^.^ , and no i cant exactly place my finger on the property.

Answer 18

sorry that took so long my ISS teacher can by to give me MORE log homework

Answer 19

That's neither here nor there :P
Anyway, the property in question is this :
\[\huge \color{blue}b^{\log_\color{blue}b\color{red}x}=\color{red}x\]
Namely, that when a number is raised to the logarithm of x with base THAT number, it simplifies to x.

Answer 20

So, how does that property figure into this :
\[\LARGE 2^{\log_2(15)}=\color{green}?\]

Answer 21

er, when you say That number simplifies into x would you mean to say the (2) would simplifie into the 15?

Answer 22

I take it you want to say this:
\[\LARGE 2^{\log_2(15)}= \color{green}{15}\]
\[\Large ?\]

Answer 23

yes precisely

Answer 24

but then again i have a feeling that it is wrong.

Answer 25

You are correct.
Are you afraid of making mistakes? Mistakes are inevitable (due to the inconvenience of being human)
Though of course, now, of all times, is the time to make mistakes, while correction is still feasible... don't worry about erring, okay, it's part of life :P

Back to reality :
\[\Large 2^a = 2^{\log_2(15)}\]Then simplifies into
\[\Large 2^a = 15\]
correct?

Answer 26

yes because they have the same base correct?

Answer 27

2 raised to a log base 2, in that respect yes..
We have successfully gotten rid of the logarithm, but don't lose sight of our main goal, which is to express a in terms only of logarithm base 7.

So far our expression is simple, not a log in sight, so what say we reintroduce a log, only this time, a log base 7.

Since they are equal, their logarithms at base 7 should also be equal:
\[\LARGE \log_7(2^a) = \log_7(15)\]

Answer 28

Now this bit:
\[\Large \log_7(2^a)\] should remind you of another property... does it ring a bell?

Answer 29

i can honestly say she has never told us any properties for logs..

Answer 30

Such injustice... or is it?
The property in question is this ;
\[\Large \log_\color{blue}b\color{red}x^\color{green}p = \color{green}p\log_\color{blue}b\color{red}x\]
Or, the logarithm of x raised to some power is equal to the product of that power and the logarithm of x (no power)

Answer 31

That said, what can be done about
\[\Large \log_7(2^a) = \color{blue}?\]

Answer 32

I feel good because i just helped someone out with an equation, but on a serious note I'm trying to figure out how \[\log_{b}x^p fits into \log_{7}(2a) \]

Answer 33

on the right hand side of that equation what does the P stand for?

Answer 34

the power to which x is raised...
and.. it's not
\[\Large \log_7(2a)\] by the way, it's \[\Large \log_7(2^a)\] a is an exponent...

Answer 35

ah i diddnt know how to raise it anyways i am sorry i am being so slow but this stuff is hard lol, what would A be then?

Answer 36

a was originally \[\Large \log_2(15)\] if you recall -_-


We're just trying to re-express it in terms of logs base 7.

So... anyway, just treat it as a variable.