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\[\log_7M=\frac{\log_910-\log_95}{\log_97}\]
first use this log law in the numerator of the right hand side \[\boxed{\log_b x-\log_by=\log_b\frac xy}\]

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Other answers:

yeah i did that so i got log 2
now use the change of base formula \[\large\boxed{\log_b x=\frac{\log_c x}{ \log_c b}}\]
0.31 for the numerator
leave it in log form
ok
So you have \[\log_7M=\frac{\log_92}{\log_97}\] comparing with the change of base formula \[\boxed{\log_b x=\dfrac{\log_c x}{ \log_c b}}\] we see that \(M= ~ . . .\)
so will u get Log7M= 0.35 @UnkleRhaukus
so will M =2 ?????
dont convert to decimal
yeah that's it M=2

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