Question

how to solve: 3log8+9log2.
the 3 and 9 are the base numbers and these must be the same in order to add logarithms

Answer 1

\[\log_38+\log_92\]

Answer 2

ow okay, so now both bases are 10

Answer 3

use the change of base formula on both terms
\[\log_b(a) = \frac{\log_{10}(a)}{\log_{10}(b)}\]

Answer 4

So the answer is log531441 (3^8 x 9^2)

Answer 5

what ?
that dosen't look right, the answer is nice and simple

Answer 6

\[\log_38+\log_92=\frac{\log8}{\log3}+\frac{\log2}{\log9}\]

Answer 7

now add the fractions, as you would if they were natural numbers rather than logs:
\[\frac ab+\frac cd = \frac{ad+bc}{bd}\]
...

Answer 8

so \[\frac{ (\log8xlog9)+ \log3 x \log9 }{ \log3 x \log9 }\]

Answer 9

So the answer must be log72, since you can take away log3 x log9

Answer 10

yes good work,


now lets simplify some of those individual log terms

with: \[\log (a^n)=n\log(a)\]

Answer 11

wahat what did you just say?

Answer 12

take away?

Answer 13

wait i think i see a mistake

Answer 14

the is a log3 x log 9 at the top and the bottom, because of that you can leave them away

Answer 15

\[\log_38+\log_92=\frac{\log8}{\log3}+\frac{\log2}{\log9}\\
=\frac{ \log8\cdot\log9+ \log3 \cdot \log2 }{ \log3 x \log9 }\]

Answer 16

ow it must be ac+bc/bd

Answer 17

\[\frac AB+\frac CD = \frac{AD}{BD}+\frac{BC}{BD}=\frac{AD+BC}{BD}\]

Answer 18

okay, now I get that. But how can it be simplified?

Answer 19

can you express 8 like a^n

Answer 20

Can't you leave the log9 and log 3 away both at the top and bottom so you are left with log8 . log2 ?

Answer 21

8 is 2^3

Answer 22

so log 8 = log 2^3 = 3 x log2


what about

log 9 =

?

Answer 23

2. log 3

Answer 24

We have:
\[\log_38+\log_92=\frac{\log8}{\log3}+\frac{\log2}{\log9}\\
=\frac{ \log8\cdot\log9+ \log3 \cdot \log2 }{ \log3 \cdot \log9}\\
=\frac{ (3\times \log2)\cdot(2\times\log3)+ \log3 \cdot \log2 }{ \log3 \cdot( 2\times\log3)}\]

Answer 25

yes, that's clear

Answer 26

now simplify the numerator

Answer 27

\[\frac{ (4 \times \log2)+(3\times \log3 }{\log3 . (2\times \log 3) }\]

Answer 28

and now you have log2^4+log3^3 at the numerator

Answer 29

um, not really

\[(3\times\log2)\cdot(2\times\log3)=(3\times2)\times \log2\cdot\log3\]


so the numerator is
\[(3\times2)\times\log2\cdot\log3+\log3\cdot\log2\\
=(3\times2+1)\log_2\log_3\]

Answer 30

Oh okay I get it now

Answer 31

typo last line should be^*\[=(3\times2+1)\log(2)\log(3)\]

Answer 32

So (6 x )log2 x log 3

Answer 33

sorry 7

Answer 34

yesh, numerator is 7 log(2) log(3)

Answer 35

now lets simplify the denominator

\[\log3\cdot(2\times\log3)\]

use \(\log_b^2(a)\), etc for \(\log_b(a)\times\log_b(a)\)

Answer 36

log3 . (log3^2) ?

Answer 37

we want to keep the 2 out the front

Answer 38

owh so 2 x (log9) ?

Answer 39

No that's wrong

Answer 40

log 3 + log 3 is log 9

log 3 x log 3 is not log 9


log 3 x log 3 is log^2 3

Answer 41

so what does our fraction look like now?

Answer 42

\[\frac{ 7 \times (2 \times \log3) }{ 2 \times \log 8 }\]

Answer 43

try again

Answer 44

\[\frac{ 7 \times (\log2 \times \log3) }{ 2 \times \log3 + \log2^3 }\]

Answer 45

\[\frac{7\log(2)\log(3)}{2\log^2(3)}=\frac{7\log(2)\log(3)}{2\log(3)\log(3)}\]

Answer 46

ow because you have log x log= log^2

Answer 47

don't mix up \[\log(a)+\log(a)=\log(a^2)\]with \[\log(a)\times\log(a)=\log^2(a)\]

Answer 48

okay didn't know that's possible with log. Clear now

Answer 49

now we can cancel the common term in the fraction...

Answer 50

so: \[\frac{ 7 \times \log2 }{ 2 \times \log3 }\]

Answer 51

canceled log 3 at the top and bottom

Answer 52

yes,

Answer 53

now you could leave it like that, with logs to base 10,

or use the change of base formula in reverse ...


\[\frac{\log(a)}{\log(b)}=\log_b(a)\]

Answer 54

how you change the base formula in reserve since you're left with the 7 and 2 ?

Answer 55

take the 7/2 out the front, use the change of base only on log(2) / log(3)

Answer 56

okay so 7/2 ( 2log3)

Answer 57

\[\frac72\log_3(2)\]

ie
(7/2) log_3 (2)

Answer 58

okay thanks alot :)

Answer 59

logs are tricky