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Solve for x by using base 10 logs.
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+11^x+2 = 12^7x
Is this what you mean?: \[11^x+2=12^{7x}\]
Or\[11^x+2=12^7x\]
or \[11^{x+2}=12^{7x}\]
yea but 11^(x+2) and the first one 12^(7X)
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yes
that last one
Okay (equation editor is for your use) \[11^{x+2}=12^{7x}\] \[\log_{10}{11^{x+2}}=\log_{10}12^{7x}\] \[\log_{10}{11^{x+2}}=\log_{10}12^{7x}\] \[(x+2)\log_{10}{11}=7x\log_{10}12\] \[(x+2)\log_{10}{11}=7x\log_{10}12\] \[2\log_{10}{11}=x(7\log_{10}12-\log_{10}{11})\] \[\frac{2\log_{10}{11}}{(7\log_{10}12-\log_{10}{11})}=x\] Does that look right?
yep sounds right thnks :)
I don't know why it repeated itself there, but okay.
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