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Would someone explain how to set this problem up? Solve the equation: log(8)x + log(11)x = 1
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8 and 11 are in base?
yes
\[\Large \log_8 x + \log_{11} x =1\] \[\frac{\log x}{\log 8} + \frac {\log x}{\log 11} = 1\]
thank you
so then what steps would allow me to solve for x ?
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take lcm.
88 ?
\[\frac{ (\log x * \log11) + (\log x * \log 8)}{\log8 * \log11}=1\]
ok, ty!
so then x= 8, 11
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Nope.
\[\frac {\log x (\log 11 + \log8)}{\log 88} = 1\]
ok, my problem asks: x= or there is no solution.
Solve for x. \[\log x = \frac{\log 88}{\log11 + \log8}\]
Insert the values in a calc.
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