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- anonymous

How can I simplify this:
log10(4)log10(2)
Please help

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- anonymous

How can I simplify this:
log10(4)log10(2)
Please help

- schrodinger

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- anonymous

log(ab) = log a + log b
so
log 10(4) log 10(2) = (log10 + log4)(log 10+log2)
log 10 = 1
(log10 + log4)(log 10+log2)=(1+log4)(1+log2)

- anonymous

I dont get it. But the original expression was:
\[Log _{16}(a)+Log _{4}(a)+Log _{2}(a)\]

- anonymous

I was trying to change the log to a unique base. and I got:
\[\log(a)=7Log(16)Log(4)Log(2)/(\log4Log2+Log16*Log2+Log16*Log4)\]
I got stacked there

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- anonymous

are you asked to find log(a)?

- anonymous

Im asked to find a

- anonymous

i think the expression isn't complete
Log16(a)+Log4(a)+Log2(a)
it should be an equation or you can't find a

- anonymous

Yeah I was thinking about that too. Ok thanks

- anonymous

Sorry the expresion was:
Log16(a)+Log4(a)+Log2(a) =7

- anonymous

do you think the equation can be solved now?

- anonymous

yes

- anonymous

can you explain it to me. I tried really hard and I got no much further.

- anonymous

log16a.4a.2a = 7
log(2^7)(a^3) = 7
7 equals to log 10^7 so
log (2^7)(a^3) = log 10^7
(2^7)(a^3) = 10^7
a^3 = 10^7 / 2^7 = (10/2)^7
a^3 = 5^7
a = 5^(7/3)

- anonymous

i got 2^7 from 16x4x2
16 is 2^4 and 4 is 2^2
therefore 16x4x2 = 2^7

- anonymous

what do you mean by:
Log16a.4a.2a = 7
\[Log(16a*4a*2a)\]
Is 10 the base? How did you end up with that?

- anonymous

yes , if the base isn't written, it means the base is 10.
one of the identities of logarithm is
loga + logb = log(ab)

- anonymous

Remember the original expression has different bases.
\[Log _{16}(a)+Log _{4}(a)+Log _{2}(a)=7\]

- anonymous

oh sorry, i thought the 16 is inside the logarithm.
so the base is 16, 4 and 2?
wait a moment i'll try solve it again

- anonymous

Ok no problem

- anonymous

Thanks

- anonymous

ok i got it
16 = 2^4
and 4\[a ^{c}\]= 2^2
so
\[\log _{2^4}a + \log _{2^2}a + \log _{2}a = 7\]
the identity of logarithm:
\[\log _{a^b}c = (1/b)logc\]
\[(1/4)\log _{2}a + (1/2)\log _{2}a + \log _{2}a = 7\]
\[(7/4)\log _{2}a = 7\]
\[\log _{2}a = (4/7) 7 = 4\]
we know that if \[\log _{a}b = c \]
then \[a ^{c}=b\]
\[\log _{2}a = (4/7) 7 = 4\]
therefore \[2^{4} = a\]
a=16

- anonymous

dont mind anything i type above the sentence " the identity of logarithm", it's typo

- anonymous

yeah i was thinking about that. Let me check it out and try to understand it. Thanks dude

- anonymous

you're welcome

- anonymous

\[\log_{a ^{b}}x ^{y}=(y/b)\log_{a}x \]
put 16 and 4 as a power of 2..and use the above log property. hope this would help

- anonymous

yeah thanks. It is clear now

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