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PDF Editor FAQ

If [math]a+b=3[/math] and [math]a^2 +b^2=7[/math], what is [math]a^4+b^4[/math]?

[math]a^4+b^4[/math] begs us to consider [math](a+b)^4[/math]. Why? Because [math](a+b)^4[/math], expanded, is going to contain [math]a^4+b^4[/math], and at the same time it’s a known quantity because we got told what [math]a+b[/math] is.So, trusty binomial theorem:[math](a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4 ab^3 + b^4[/math]We’d like to rearrange this into something that plays nice with [math]a+b[/math] and [math]a^2+b^2[/math]. One natural thing comes to mind:[math](a+b)^4 = a^4 + b^4 + 4ab(a^2+b^2) + 6(ab)^2[/math]We simply collected the terms in symmetric pairs (first and last, second and next-to-last, leave the center alone.)Almost everything here is either a known quantity or the thing we’re trying to find. The only exception is [math]ab[/math]. Can we determine [math]ab[/math]?Why, sure, using the exact same trick.[math](a+b)^2 = a^2+b^2+2ab[/math][math]9 = 7 + 2ab[/math][math]ab=1[/math]Hooray, nice and simple. So back to our equation:[math]81 = a^4+b^4 + 4\cdot 7 + 6[/math]and we’ve determined that [math]a^4+b^4[/math] must be [math]81-(4 \cdot 7 + 6) = 47[/math]. [math]\blacksquare[/math]

How do you interchange the value of two variables without using any other variable?

Let a and b are the two variables, thena = a + bb = a - ba = a - bThe catch is you have to look at '=' as an assignment operator (which it is :0)For eg: a = 4, b = 6a = a + b => 4 + 6 = 10now a = 10(assigned in the above step), b = 6 (of course!)b = a - b => 10 - 6 = 4now a = 10(nothing changed, no assignments to a ), b = 4a = a - b => 10 - 4 = 6Which gives a = 6, b = 4 ( Hence Interchanged )

If [math]3^a= 4 , 4^b= 5 , 5^c= 6[/math] , [math]6^d= 7[/math], then what is the value of [math]abcd[/math]?

[math]\displaystyle 3^{abcd}=(3^a)^{bcd}=4^{bcd}=(4^b)^{cd}=5^{cd}=(5^c)^d=6^d=7[/math].So [math]abcd[/math] is a number which makes [math]3^{abcd}[/math] equal [math]7[/math]. Among real numbers, the common name for this number is [math]abcd=\log_3(7)[/math].

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