Question

How can I simplify this:

log10(4)log10(2)

Please help

Answer 1

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)

Answer 2

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

Answer 3

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

Answer 4

are you asked to find log(a)?

Answer 5

Im asked to find a

Answer 6

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

Answer 7

Yeah I was thinking about that too. Ok thanks

Answer 8

Sorry the expresion was:

Log16(a)+Log4(a)+Log2(a) =7

Answer 9

do you think the equation can be solved now?

Answer 10

yes

Answer 11

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

Answer 12

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)

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

Remember the original expression has different bases.

\[Log _{16}(a)+Log _{4}(a)+Log _{2}(a)=7\]

Answer 17

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

Answer 18

Ok no problem

Answer 19

Thanks

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

you're welcome

Answer 24

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

Answer 25

yeah thanks. It is clear now