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Solve 8^x – 2 = 3
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start with \(8^x=5\)and then recall that \[b^x=A\iff x=\frac{\ln(A)}{\ln(b)}\]
Log8^x =log5 => x =log5/log8 =log(5-8)
unless of course it is \(8^{x-2}=3\) in which case you get \[x-2=\frac{\ln(3)}{\ln(8)}\] and so \[x=\frac{\ln(3)}{\ln(8)}+2\]
careful careful here @sachinrajsharma it is not true that \(\frac{\log(a)}{\log(b)}=\log(a-b)\)
is it 2.52?
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depends on the problem
its the second one you wrote out
is it \[8^{x-2}=3\] or is it \[8^x-2=3\]?
the first one
ok so \[8^{x-2}=3\] right? yea, 2.528...
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if you are going to round, use 2.53
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