Question

help needed

Answer 1

\[\log_7M=\frac{\log_910-\log_95}{\log_97}\]

Answer 2

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

Answer 3

yeah i did that so i got log 2

Answer 4

now use the change of base formula
\[\large\boxed{\log_b x=\frac{\log_c x}{ \log_c b}}\]

Answer 5

0.31 for the numerator

Answer 6

leave it in log form

Answer 7

ok

Answer 8

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 \(M= ~ . . .\)

Answer 9

so will u get Log7M= 0.35 @UnkleRhaukus

Answer 10

so will M =2 ?????

Answer 11

dont convert to decimal

Answer 12

yeah that's it M=2