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log2(x-3)=5
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\[\log_{2} (x-3)=5\]
X=35
how do you get that?
Let (x-3) = a \[\large \log_3 a = 5\] \[\large Log_a b = c ---> a^c = b\]
So, what do you get off of that?
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Try and do the work, just getting the answers won't help you.
i just needed help, not the answer
@dalia_lam Is my answer wrong than?
@dalia_lam that's incorrect \[\large \log_b M - \log _b N = \log_b \frac{ M }{ N }\]
not log_b(m-n)...
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Look at my example.. try and solve it
wait, i forgot the bracket
\[\Huge Let (x-3) = a\] \[\Huge \log_a b = c -> a^c = b \]
|dw:1386296360298:dw|
\[\Huge a = 3^5\]
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\[\Huge \log_2 a = 5\] \[\Huge a = 2^5\]
\[\Huge x -3 = 32\] \[\Huge x = 35\]
i got 35, and thats on my answer list so im guessing thats right. thank you guys for the help.
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