Question

solve

logbase5 (25x)

its not 1/25 i tried it

Answer 1

2

Answer 2

Just remember this
\[\log_a b^c = clog_a b\]

Answer 3

if you have
\[\log_5(25x)\] there is nothing to solve for. you would rewrite it as
\[\log_5(25)+\log_5(x)=2+\log_5(x)\]

Answer 4

unless it is
\[\log_5(25^x)\] which is a different story

Answer 5

@satellite73 ,Exactly, i just wanted to say that :D

Answer 6

It was the 2+logbase5(x)

Answer 7

can you help me do logbase3(9x)?

Answer 8

exactly the same idea

Answer 9

I see that you do logbase2 3+ logbase2 (x)= ?

Answer 10

use
\[\log(ab)=\log(a)+\log(b)\] to rewrite as
\[\log_3(9)+\log_3(x)\]

Answer 11

then note that
\[\log_3(9)=2\] because \[3^2=9\]

Answer 12

ok i got that

Answer 13

so answer would be
\[2+\log_3(x)\]

Answer 14

ohhhhhhh ok thank you i was wondering where the 2 came from but now i see

Answer 15

oh good. same place it came from in the first one right?
\[\log_5(25)=2\] because
\[5^2=25\]

Answer 16

yes [=