Question

Need help with the following:

Evaluate exactly:

log9(729) + 3log32(2) - log(100,000,000) + 2ln(e^12)

Answer 1

what would you get from say \(\bf 9^3\quad ?\)

Answer 2

729

Answer 3

how bout the second term would that become a 6?

Answer 4

well let us keep in mind that so
\(\large \bf log_{\color{red}{ a}}{\color{blue}{ b}}=y\implies {\color{red}{ a}}^y={\color{blue}{ b}}
\\ \quad \\ \quad \\

log_9(729) + 3log_{32}(2) - log(100,000,000) + 2ln(e^{12})
\\ \quad \\
\implies log_9(729) + 3log_{32}(2) - log_{10}(100,000,000) + 2log_e(e^{12})
\\ \quad \\
{\color{blue}{ 729\implies 9^3\qquad 32\implies 2^5\qquad 100000000\implies 10^8}}\qquad thus
\\ \quad \\
log_9(9^3) + 3log_{2^5}(2) - log_{10}(10^8) + 2log_e(e^{12})
\\ \quad \\
log_9(9^3) + log_{2^5}(2^{\color{blue}{ 3}}) - log_{10}(10^8) + log_e(e^{12\cdot {\color{blue}{ 2}}}) \)

Answer 5

the 1st, 3rd and last term are simple, if you notice -> \(\bf \Large log_{\color{red}{ a}}{\color{blue}{ b}}=y\implies {\color{red}{ a}}^y={\color{blue}{ b}}\) and change them to exponential notationn

Answer 6

how bout the second and last one? i don't get how you got what those are

Answer 7

do you see what the 1st and 3rd are?

Answer 8

so thus far we have \(\bf \Large { log_9(9^3) + log_{2^5}(2^{\color{blue}{ 3}}) - log_{10}(10^8) + log_e(e^{12\cdot {\color{blue}{ 2}}})
\\ \quad \\
log_9(9^3) + log_{2^5}(2^{3}) - log_{10}(10^8) + log_e(e^{24})
}\)

Answer 9

ok right

Answer 10

anything confusing yet?

Answer 11

the part where you made 3log32(2) log2^5(2^3)
but i see what you did

Answer 12

so now we can use the laws of logs right?

Answer 13

http://www.chilimath.com/algebra/advanced/log/images/rules%20of%20exponents.gif <--- notice 3rd rule listed there, thus

Answer 14

well, lemme put them in exponential form

Answer 15

ok

Answer 16

\(\bf log_9(9^3) + log_{2^5}(2^{3}) - log_{10}(10^8) + log_e(e^{24})
\\ \quad \\

log_{\color{red}{ 9}}(9^3)=\square \implies {\color{red}{ 9}}^\square =9^3\implies \square =3\qquad thus log_{\color{red}{ 9}}(9^3)=3
\\ \quad \\
{\large log_{{\color{red}{ 2^5}}}(2^3)} =\square \implies ({{\color{red}{ 2^5}}})^\square =2^3\implies {{\color{red}{ 2^{5\cdot \square }}}}=2^3\implies 5\cdot \square =3\)

can you tell what \(\square\) is from there?

Answer 17

\(\bf log_9(9^3) + log_{2^5}(2^{3}) - log_{10}(10^8) + log_e(e^{24})
\\ \quad \\

log_{\color{red}{ 9}}(9^3)=\square \implies {\color{red}{ 9}}^\square =9^3\implies \square =3\qquad thus \quad log_{\color{red}{ 9}}(9^3)=3
\\ \quad \\
{\large log_{{\color{red}{ 2^5}}}(2^3)} =\square \implies ({{\color{red}{ 2^5}}})^\square =2^3\implies {{\color{red}{ 2^{5\cdot \square }}}}=2^3\implies 5\cdot \square =3\)

Answer 18

so... what do you think will be for \(\square\quad ?\)

Answer 19

not sure what the box is

Answer 20

well. how about if you just solve fro \(\square\) ? :)

Answer 21

if you don't like \(\square\) you can call it "x" if you want
thus 5 * x = 3 = > 5x =3 solve for 'x"

Answer 22

ok so ive been staring at this and it helps to solve for x

so far i think that the first term log9(9^3) = 3 if im not mistaken
is that correct?

Answer 23

the second one is then 3/5

Answer 24

is the answer -22/5?

Answer 25

hmmm dunno

Answer 26

have you gotten the 3rd and last term yet?

Answer 27

and yes \(\bf {\large log_{{\color{red}{ 2^5}}}(2^3)} =\square \implies ({{\color{red}{ 2^5}}})^\square =2^3\implies {{\color{red}{ 2^{5\cdot \square }}}}=2^3\implies 5\cdot \square =3
\\ \quad \\
\large \square =\cfrac{3}{5}\qquad thus\qquad log_{{\color{red}{ 2^5}}}(2^3)=\cfrac{3}{5}\)

Answer 28

recaalll that all we're doing is chainging to exponential notation, that is \(\Large \bf log_{\color{red}{ a}}{\color{blue}{ b}}=y\implies {\color{red}{ a}}^y={\color{blue}{ b}}\)

Answer 29

sorry i meant -12/5

and yeah i got all the terms

3 + (3/5) - 8 + 2

i got -12/5

Answer 30

thank you jdoe001 you really helped me out alot

Answer 31

close.... one sec let us get the 3rd and 4th term

Answer 32

ok

Answer 33

\(\bf log_9(9^3) + log_{2^5}(2^{3}) - log_{10}(10^8) + log_e(e^{24})
\\ \quad \\
log_{{\color{red}{ 10}}}(10^8)=\square \implies {{\color{red}{ 10}}}^\square =10^8\implies \square =8\qquad thus\quad log_{{\color{red}{ 10}}}(10^8)=8
\\ \quad \\
log_{\color{red}{ e}}(e^{24})=\square \implies {\color{red}{ e}}^\square =e^{24}\implies \square =24\qquad thus\quad log_{\color{red}{ e}}(e^{24})=24\)

Answer 34

ok i didnt realize the e wasn't being raised to any powers and i got 2 when i solved for the last log

Answer 35

recall the coefficient in front of it -> \(\bf \large {log_9(9^3) + 3log_{2^5}(2) - log_{10}(10^8) + 2log_e(e^{12})
\\ \quad \\
log_9(9^3) + log_{2^5}(2^{\color{blue}{ 3}}) - log_{10}(10^8) + log_e(e^{12\cdot {\color{blue}{ 2}}}) }\)

Answer 36

ok 2nd and 4th are done the same way

Answer 37

ahemm I thought we did those already =)

Answer 38

but yes, 2nd and 4th of them are done the same way

recall -> http://www.chilimath.com/algebra/advanced/log/images/rules%20of%20exponents.gif rule 3 listed there

Answer 39

yeah we did let me do the whole thing with everything we just went over and ill post the result and i think i should get it right this time

Answer 40

ok

Answer 41

\(\bf log_{\color{red}{ a}}{\color{blue}{ b}}=y\implies {\color{red}{ a}}^y={\color{blue}{ b}}
\\ \quad \\ \quad \\
log_9(9^3) + log_{2^5}(2^{3}) - log_{10}(10^8) + log_e(e^{24})
\\ \quad \\
------------------------------------\\
log_{\color{red}{ 9}}(9^3)=\square \implies {\color{red}{ 9}}^\square =9^3\implies \square =3\qquad thus \quad log_{\color{red}{ 9}}(9^3)=3
\\ \quad \\
------------------------------------\\
{\large log_{{\color{red}{ 2^5}}}(2^3)} =\square \implies ({{\color{red}{ 2^5}}})^\square =2^3\implies {{\color{red}{ 2^{5\cdot \square }}}}=2^3\implies 5\cdot \square =3
\\ \quad \\
\large \square =\cfrac{3}{5}\qquad thus\qquad log_{{\color{red}{ 2^5}}}(2^3)=\cfrac{3}{5}
\\ \quad \\
------------------------------------\\

log_{{\color{red}{ 10}}}(10^8)=\square \implies {{\color{red}{ 10}}}^\square =10^8\implies \square =8\qquad thus\quad log_{{\color{red}{ 10}}}(10^8)=8
\\ \quad \\
------------------------------------\\
log_{\color{red}{ e}}(e^{24})=\square \implies {\color{red}{ e}}^\square =e^{24}\implies \square =24\qquad thus\quad log_{\color{red}{ e}}(e^{24})=24\)

Answer 42

answer is 19.6