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

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

Mathematics
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so - note that 45 = 9*5 you have loga 5 and oyu have log a 3 what is the relationship betweem 3 and 9?
im not sure
2^2 = 4 3^2 = ?

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9
SO 9 = 3^2 So if loga3 = 1.59 what is loga 3^2
would it be 3
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 \[\log _{a} x ^{n}\]
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?
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
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?
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
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
if it's not too much to ask, could you also help me with this last one? i have to condense \[\log_{4} 128-1\]
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
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...)
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?
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?
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
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....
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!!

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