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anonymous
 3 years ago
Another log problem...
anonymous
 3 years ago
Another log problem...

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perl
 3 years ago
Best ResponseYou've already chosen the best response.0may i ask how you uploaded that attachment, thanks how did you put it in an attachment

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got it to be (x+6)(x+4)=8 But the answer is 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@perl you can attach a file.. and i had a picture of the equation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0close you're on the right track. remember the priciples \[\log_{z} b+\log_{z} c=\log_{z} bc\] and if \[\log_{a}b=c \] then \[a^{c}=b\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0pls tell me the steps that you have followed...so that i can spot the mistake?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0log8(x + 6) = 1 – log8(x + 4) log8(x + 6) + log8(x + 4) = 1 Using the properties of logarithms: log A + log B = log (AB) log8[(x + 6) (x + 4)] = 1 Rewrite in exponential form: 8¹ = (x + 6) (x + 4) x² + 10x + 24 = 8 x² + 10x + 16 = 0 (x + 2) (x + 8) = 0 x = 2 or x = 8 x = 8 is an extraneous solution, because8 + 6 = 2, and 8 + 4 = 4, and log base 8 is undefined for negative numbers. (The domain is positive numbers only.) So the solution is x = 2. Verify: log8(4) = 1 – log8(2) log8(4) is 2/3 (Because 8^(2/3) = 2² = 4) log8(2) is 1/3 (Because 8^(1/3) = 2) So the above equation is equivalent to: 2/3 = 1 – 1/3 2/3=2/3
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