Question

1 - 2log7^x (7 is the base)

Answer 1

Is this even possible? to subtract a log from a number?

Answer 2

1 can be written as a log

Log a to the base a is = 1

Answer 3

\[\log_{a} a = 1\]

Answer 4

here your base is 7 . So express it as log and continue like we did the last problem

Answer 5

ok

Answer 6

can you show me your expression after you wrote 1 as a log

Answer 7

I'm still kinda confused on changing 1 into a log

Answer 8

oh waitttt

Answer 9

\[\log_{a} a = 1\]

Answer 10

here we need to have a base of 7

Answer 11

ok

Answer 12

so log7 and then 1/x? I'm sure that's wrong.

Answer 13

\[\log_{7} 7 = 1\]

why 1/x, what is that?

Answer 14

now we have\[\log_{7} 7 - 2\log_{7} X\] You write them as a single log like we did our earlier problem

Answer 15

and it'll come out as a fraction, right?

Answer 16

log then fraction

Answer 17

ok so log7 7/x^2 ?

Answer 18

Fantastic \[\log_{7} 7 - 2 \log_{7} X = \log_{7} 7 - \log_{7} X ^{2}\]\[= \log_{7} \frac{ 7 }{ X ^{2} }\]

Answer 19

yeahhhhhhh!!!!!

Answer 20

Thanks again.