how do you solve this using logs?

The change in base from a to b is

log_b(x) = log_a(x)/log_a(b)

So if b = 4 and a = 2

log_4(9) = log_2(9)/log_2(4)
log_4(9) = log_2(9)/2
2log_4(9) = log_2(9)

2log_4(9) - log_2(3) = log_2(9) - log_2(3)
2log_4(9) - log_2(3) = log_2(9/3)

2log_4(9) - log_2(3) = log_2(3)

=2log(base4)9 - log(base2)3

=( 2log(base2)9 / log(base2)4 ) - log(base2)3

=( 2log(base2)9 / 2 ) - log(base2)3

=log(base2)9 - log(base2)3

=log(base2)3

2 log 9/log4-log 3/log 2=
=2log 9/ 2log 2 -log 3 / log 2=
=log 9/ log 2 - log 3/log 2=
=log 3/log 2=log(base 2) 3
God bless you.

2log(base 4)9 - log(base2)3

Use change of base with base 10, that way you can evaluate on your calculator.

2[(log9)/(log4)] - [(log3)/(log2)]
2[1.585] -[1.585] -------> notice ((log9)/(log4)) = ((log3)/(log2))
1.585

You can't!!!

log2 4 + log3 9
por favor no encuentro la solución
gracias

2log_4(9) - log_2(3)
= 2log_(2^2)(9) - log_2(3)
= 2[log_2(9)]/2 - log_2(3)
= log_2(9) - log_2(3)
= log_2(9/3)
= log_2(3)

log(base a)x = log(base a)b .log(base b)x
so
log(base 4)9 = log(base 4)2. log(base 2)9 = .(1/2)log(base 2)9 = log(base 2)9^(1/2) = log(base 2)3
so
2log(base 4)9 = 2log(base 2)3 = log(base 2)9
Therfore
2log(base 4)9 - log(base 2)3 = log(base 2)9 - log(base 2)3
= log(base 2)(9/3)
= log(base 2)3

2log4(9)-log2(3)
Base change says that you can change the log2(3) to log4(3)/log4(2) and log4(2) is just 1/2. So it's the same as 2log4(3)
Now you hae
2log4(9)-2log4(3)
=2log4(9)-log4(3^2)
=2log4(9)-log4(9)
=log4(9)
that's about 1.584

first, let [] denote the base:

a * log[b](c) = log[b](c^a)

also:

log[b](c) = log(c) / log(b)

So:

log(81) / log(4) - log(3) / log(2)

log(4) = log(2^2) = 2 * log(2)

log(81) = log(3^4) = 4 * log(3)

4 * log(3) / (2 * log(2) - log(3) / log(2)
2 log(3) / log(2) - log(3) / log(2)
log(3) / log(2)
log[2](3)