question in comments (logarithmic question)

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question in comments (logarithmic question)

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[\log _{4}(8-x)=3\] a) 56 b) -56 c) -73 d) 4 e) 72
convert log to exponential form example |dw:1437344474349:dw|
@Nnesha thanks! got it.

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are you sure?? let me know plz
@Nnesha could you help me with some other logarithmic questions? #1: ln(x)+7=12 a) e^(12)-7 b)5^e c) e^(7)-12 d) 5 e) e^5 #2: \[\log _{2}(x)+\log _{2}(x-4)=\log _{2}(12)\] a) x=-2, x=6 b) x=0, x=4 c) x=2, x=-6 d) x=6 e) x=-2
alright okay so first one how would you cancel out ln?? do you know?
no, can you explain it to me?
okay to cancel out ln you have to take e both side example \[\huge\rm \cancel{e^{\ln}} x = x\]
for example in this question i have to solve for x ln x = 3 so i would take ln both sides \[\huge\rm \color{red}{e}^{\ln} x= \color{ReD}{e}^3\] \[\huge\rm \cancel {\color{red}{e}^{\ln}} x= \color{ReD}{e}^3\] answer is x = e^3
@Nnesha awesome! now could you help me with the last one?
you have to apply log property there
what's log property?
mhm you can start log questions without knowing property! :)quotient rule\[\huge\rm log_b y - \log_b x = \log_b \frac{ x }{ y}\] to condense you can change subtraction to division product rule \[\huge\rm log_b x + \log_b y = \log_b( x \times y )\] addition ----> multiplication
okay so once i have \[\log _{2}(x ^{2}-4x)=\log _{2}(12)\] what do I do?
nice so there are same bases both sides you can cancel each other ot \[\huge\rm \color{ReD}{log_b }x = \color{red}{\log_b} y \] \[\huge\rm\cancel{ \color{ReD}{log_b }}x = \cancel{\color{red}{\log_b}} y \] x=y
@Nnesha got it! thanks so much for all the help!
my pleasure

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