Question

Solve
log6x^8-log6x^5=31

Answer 1

at first use \(\Large\color{black}{ \bf log(a^{b})=b~log(a) }\)

Answer 2

re-write using the above rule

\(\Large\color{black}{1)~~~~ \bf log~(6x~^{8})~=~? }\)
\(\Large\color{black}{2)~~~~ \bf log~(6x~^{5})~=~? }\)

the re-write the problem again.

Answer 3

then re-write...

Answer 4

Is it this? \(log_{6}\left(x^{8}\right)\)

Answer 5

yes , well the x^8 wasn't in "( )" but yes

Answer 6

\(\Large\color{black}{\bf log_6x^8-log_6x^5=31 }\)

Answer 7

@SolomonZelman wouldn't it rewrite to \[8 \log_{6} (x)-5\log_{g}(x) =31\]

Answer 8

YES! \(\Large\color{black}{\bf 8~log_6x-5~log_6x=31 }\)

Answer 9

Which simplifies on the left side to,

\(\Large\color{black}{\bf 3~log_6x=31 }\)

Answer 10

can you finish the problem or need more help?

Answer 11

then you divide both sides by 3 correct?

Answer 12

I am lazy and have no calc....

\(\Large\color{black}{\bf 3log_6x=31 \times log_66 }\)

\(\Large\color{black}{\bf log_6x^{3}= log_66^{31} }\)

\(\Large\color{black}{\bf x^{3}= 6^{31} }\)

Answer 13

to get x=\[6 ^{10\frac{ 1 }{ 3 }}\]

Answer 14

See what I did in my last post ?

Answer 15

yeah, but i don't seem to understand how you got that.If you don't mind could explain that please?

Answer 16

\(\log^{b}(a) = c \iff b^{c} = a\)

Answer 17

Yes, I multiplied the left side times magic 1, (which is log(base 6) ( 6 ) )

Answer 18

I multiplied the right side, not the left side, ... sorry.

Answer 19

ohh i understand now thanks

Answer 20

You welcome !

Answer 21

If you don't mind could you help me with one more problem please?

Answer 22

\(\log_{b}(a) = c\iff b^{c} = a\)

Not sure how that b managed to get up in the attic.