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Solve for x when.. log(base10)x+log(base10)(x-3)=1
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\[\log_{10}x+\log_{10}(x-3)=1 \]
First, write the two logs on the left as a single log using the product rule for logs
so \[\log_{10} x(x-3)=1\] ??
Yah!!! good. So now rewrite the log equation as an exponential equation. Hint: \(\large \log_b(x) = y\) if and only if \(\large b^y = x\)
\[x ^{2}-3x-1=0\]x
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almost. that 1 should be 10, \[\large x(x-3) = 10^1\]
ooooooooo okay see I was going to do that but wasn't sure if that seemed right. So then from there factor it out and solve for x?
You got it!
Ooh wow it's actually not that hard Thank you so much!
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