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How can I using algebra, solve for this, thank you

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by camitava » Mon Nov 05, 2007 8:46 pm
Skyjuice, we can write -
(2^x)(3^y)=288 in (2^x)(3^y)=2^5 * 3^2
So x = 5 and y =2
And the rest thing we can calculate, right!
Correct me If I am wrong


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by samirpandeyit62 » Mon Nov 05, 2007 8:50 pm
If (2^x)(3^y)=288 where x and ya are positive intergers then (2^x-1)(3^y-2)= ?

(2^x)(3^y)=288

now (2^x-1)(3^y-2)= (2^x)(3^y)/2*9 = 288/18 =16
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by moneyman » Mon Nov 05, 2007 9:44 pm
Samir can u explain this step pls

(2^x-1)(3^y-2)= (2^x)(3^y)/2*9 = 288/18 =16

How did u get 2*9??
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by camitava » Mon Nov 05, 2007 9:51 pm
Moneyman, Samir has taken -
(2^x-1)(3^y-2)= (2^x)(3^y)/2*9 = 288/18 =16
by this what he tried to mean that 2^x - 1 = 2^(x - 1) and same for the rest.
I am still confused what skyjuice wanted to mean it - is it 2^x - 1 or 2^(x - 1). Got my point, moneyman!
Correct me If I am wrong


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by samirpandeyit62 » Mon Nov 05, 2007 9:51 pm
(2^x-1)(3^y-2)= (2^x)(3^y)/2*9 = 288/18 =16

(2^x-1) = 2^x /2

(3^y-2)= 3^y / 3^2
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by moneyman » Mon Nov 05, 2007 9:58 pm
I think I have got the logic here

(2^x-1) = 2^x/2 because lets assume x=3 then 2^x=2^3=8 and (2^x-1)=2(3-1)=2^2=4 or 2^3/2.

Thanks a lot Samir!!
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