anonymous
  • anonymous
a) log4 (x+8) = log2 3 b) 3 ^2x= 5/3x c) log3 x = 2-log3 (x+8) d) log3 (2y-4) = 2 log3 2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[a) \log_{4} (x+8) = \log_{2} 3\]
anonymous
  • anonymous
\[b) 3^{2x} = 5 / 3^{x}\]
anonymous
  • anonymous
c) \[\log_{3} x=2-\log_{3} (x+8)\]

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More answers

anonymous
  • anonymous
\[\log_{3} (2y-4) = 2 \log_{3} 2 \]
anonymous
  • anonymous
= log3 x2 + log3 5 = log3(5x2)
anonymous
  • anonymous
@Callisto @FoolForMath @apoorvk @aroub
apoorvk
  • apoorvk
|dw:1333455216655:dw|
apoorvk
  • apoorvk
|dw:1333455333220:dw|
apoorvk
  • apoorvk
|dw:1333455437498:dw|
apoorvk
  • apoorvk
|dw:1333455558093:dw|
apoorvk
  • apoorvk
remember always to check if the values of x or y violate the domain for the argument and base of the log. base is always greater than 0 and unequal to 1, and argument is always greater 0.
anonymous
  • anonymous
|dw:1333455914279:dw|
Callisto
  • Callisto
it's 9, i think
King
  • King
x is 1
King
  • King
see over there its 1/2log something when u simplify that u get 2log something on the other side that is log9base2
King
  • King
so x+8=9 x=1
Callisto
  • Callisto
i meant the thing that look like g is 9, the answer to that question is 1 as apoorvk have written
King
  • King
oh yeah that thing is 9 and @gF if u were asking hw its 9 then read my explanantion......
anonymous
  • anonymous
\[\log_{3} (2y-4) = 2 \log_{3} 2\] \[\log_3(2y-4)=\log_3(2^2)=\log_3(4)\] \[2y-4=4\] \[2y=8\] \[y=4\]
anonymous
  • anonymous
oops solved already !
anonymous
  • anonymous
\[ \log_{4} (x+8) = \log_{2} ( 3) \] now maybe what is confusing is that \[\log_{b^n}(x)=\frac{1}{n}\log_b(x)\] so that \[\log_4(x+8)=\frac{1}{2}\log_2(x+8)\] because \(4=2^2\)
anonymous
  • anonymous
now start with \[\frac{1}{2}\log_2(x+8)=\log_2(3)\] and multiply by 2 to get \[\log_2(x+8)=2\log_2(3)\] the right hand side becomes \(2\log_2(3)=\log_2(3^2)=\log_2(9)\) because \(n\log(x)=\log(x^n)\)

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