PLEASE HELP WITH LOGs
IF loga5=2.32 and loga3=1.59, find loga45

- anonymous

PLEASE HELP WITH LOGs
IF loga5=2.32 and loga3=1.59, find loga45

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- MrNood

so - note that 45 = 9*5
you have loga 5
and oyu have log a 3
what is the relationship betweem 3 and 9?

- anonymous

im not sure

- MrNood

2^2 = 4
3^2 = ?

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- anonymous

9

- MrNood

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

- anonymous

would it be 3

- MrNood

no
this is the key point of th equestion
you must be studying logs and their properties so see if you can find a way of simplifying
\[\log _{a} x ^{n}\]

- anonymous

well that would be n times loga(x)

- MrNood

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

- anonymous

2 times loga(3)

- anonymous

loga(5*3)

- MrNood

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

- anonymous

i mean 5*9

- anonymous

so loga(5)+loga(9)

- MrNood

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

- anonymous

but the answer key says 5.50 is that wrong?

- MrNood

no
you wrote just now what log 3^2 is (it is in YOUR post above...
what is log 9?
what is log 5
add tehm together

- anonymous

i got 2?

- MrNood

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

- anonymous

oh so 2 loga (3)

- MrNood

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

- anonymous

but im typing that into my calculator and getting 2?

- MrNood

you don't need a calculator
you are given the numbers in the question
what is loga 3?
what is 2 loga 3?
what is loga 5?
soooo - what I slog a 45
you have DONE all the work - you just need to put the steps together
you are using the calculator wrong - but that doesn't matter now - you don't need the calculator

- MrNood

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

- anonymous

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

- anonymous

thank you so much!

- MrNood

np

- anonymous

if it's not too much to ask, could you also help me with this last one? i have to condense \[\log_{4} 128-1\]

- MrNood

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

- anonymous

i dont think we learned that

- MrNood

so if it was to look like I said tehn we can see that
log4 y = 1
what is y?
(think about the meaning of what loga really means) (I think you seem to have the skills to do this...)

- anonymous

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

- MrNood

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

- anonymous

1?

- MrNood

yes
so logn n =1 for any n
so we can write your original equation as
log4 128 - log 4 4
can oyu simplify from there?

- anonymous

i got 32 but the answer is 5/2

- anonymous

because i did log4 128/4

- MrNood

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

- anonymous

would you use the change of base property

- MrNood

no - sorry - I was being slack in notation - forget that
you are nearly there
go back to
log4 128 - log 4 4
can oyu simplify from there?
And try again - you made a small mistake....

- anonymous

log4 (128/4)?

- MrNood

yes

- anonymous

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

- anonymous

thanks so much!

- MrNood

not sure I follow that

- MrNood

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

- anonymous

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

- MrNood

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

- anonymous

yeah

- MrNood

well done

- anonymous

thanks so much!!

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