anonymous
  • anonymous
Solve the equation log_3(3t+14)-log_3(t)=log_3(4)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
base 3?
anonymous
  • anonymous
??
anonymous
  • anonymous
yeah

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anonymous
  • anonymous
-1.14031399559
anonymous
  • anonymous
|dw:1449725483454:dw|
whpalmer4
  • whpalmer4
Sort of glossed over a step or two which might not be so obvious to someone who is asking for help: \[\log_3 (3t+14) - \log_3(t) = \log_3(4)\]Adding logarithms is the same as taking the logarithm of the product: \[\log a + \log b = \log (a*b)\] So that allows us to do this: \[\log_3(3t+14) - \log_3(t) = \log_3(4)\]\[\log_3(3t+14) = \log_3(4) + \log_3(t)\]\[\log_3(3t+14) = \log(4*t)\]Now we raise 3 to the power of both sides to undo the logarithm: \[3^{\log_3(3t+14)} = 3^{\log_3(4t)}\]but because \[b^{\log_b x} = x\]we can simplify that to \[(3t+14) = (4t)\]\[14 = 4t-3t\]\[t=14\] If you want to test your solution on a calculator, it becomes \[\log_3(3(14)+14) - \log_3(14) = \log_3(4)\]\[3.66403-2.40217 = 1.26186\]

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