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im not sure

2^2 = 4
3^2 = ?

SO 9 = 3^2
So if loga3 = 1.59
what is loga 3^2

would it be 3

well that would be n times loga(x)

good - you have solved the question!
so what is
\[\log _{a }3^{^{2}}\]

2 times loga(3)

loga(5*3)

no - you want to find loga(5 * 3^2)

i mean 5*9

so loga(5)+loga(9)

you got it
and you have both those numbers.....
add em up!

but the answer key says 5.50 is that wrong?

i got 2?

YOU wrote above
"well that would be n times loga(x)
so what is log 9 ( it is the same as log 3^2)

oh so 2 loga (3)

yes - you already wrote that
so now oyu can work out
loga(5)+loga(9)

but im typing that into my calculator and getting 2?

anyhow - you can't use calc because you don't know what a is

oh ok i forgot about the numbers so i got 5.5 with the number

thank you so much!

np

OK
it would be good if this was
log 4 x - log 4 y
Can you make it look like that?

i dont think we learned that

do you mean y=128 or do you want it in exponential form?

no - it is not y=128
what is
\[\log_{10} 10\]
or
\[\log_{e} e\]

1?

i got 32 but the answer is 5/2

because i did log4 128/4

ok - you have not quite done the simplification properly
what is log a - log b?

would you use the change of base property

log4 (128/4)?

yes

oh i got i! you convert it to exponential so you get 5/2

thanks so much!

not sure I follow that

log4 (128/4) = log4 (32) which IS 5/2
but I need to see how you got there....

i got 4^x =32 and then you do (2^2)^x=(2^5) and 2x=5 is 5/2

OK - I see
32 = 2^5
so 4^5 /4^2

yeah

well done

thanks so much!!