Logarithm Sums

Without loss of generality, let $x < y$ . Let $\log_x y=n$ , then $\log_y x=\frac{1}{n}$ . $n+\frac{1}{n}=\frac{17}{4}\Rightarrow n^2-\frac{17}{4}n+1=0$ . Using quadratic formula, we find that $n=4$ (since $n>1$ ). So, $xy=x\cdot x^4=x^5=288\sqrt{3}=\sqrt{248832}\Rightarrow x=248832^\frac{1}{10}=\sqrt{12}$ . Then, $y=x^4=(\sqrt{12})^4=144$ . $x+y=\sqrt{12}+144=2\sqrt{3}+144$ . $a+b+c=144+2+3=149$

log x⁡〖y +log y⁡x=17/4〗 equation 1 xy=288√3 equation 2

If we let log_x⁡〖y=n〗

it follows that 〖log_y〗⁡〖x=1/n〗

substituting assumption to equation 1

then, n+1/n=17/4

solving for n 4n^2+4=17n 4n^2-17n+4=0 to find n=4 and n=1/4

if n is equal to 4 log_x⁡〖y=4〗 y=x^4

substituting value of y to equation 2 xy=288√3 x(x^4)=(32)(9)(3^(1/2) ) x^5=(2^5)(3^2)(3^(1/2)) x=((2^5 )(3^(5/2) ))^(1/5) x=〖(2)3〗^(1/2) x=2√3

then, y=x^4 y=(〖(2)(3^(1/2) )) 〗^4

y=(16)(9) y=144

then, x+y=a+b√c x+y=144+2√3 3 is not divisible by the square of any prime number, ∴a+b+c=149

Since the equations are symmetric in $x$ and $y$ , without loss of generality let $x \leq y$ . Let $p = \log_x y$ , so $p \geq 1$ . We have $p + \frac{1}{p} = \frac{17}{4}$ . Clearing denominators and simplifying, we get $4p^2 -17 p + 4 = 0$ and this factors to $(4p - 1)(p - 4) = 0$ . Since $p \geq 1$ we have $p = 4$ . Thus $\log_x y = p = 4 \Rightarrow y = x^{4}$ .

Substituting $y = x^{4}$ into $xy = 288\sqrt{3}$ , we have $x^{5} = 288\sqrt{3} = 2^{5}3^{\frac{5}{2}} \Rightarrow x = 2\sqrt{3}$ . Thus $y = \frac{288\sqrt{3}}{2\sqrt{3}} = 144$ . Therefore $x + y = 144 + 2\sqrt{3}$ . Hence $a + b + c= 144 + 2 + 3 = 149$ .

Note: The initial assumption means that we have 2 solution pairs $(x,y) = (2\sqrt{3}, 144), (144, 2 \sqrt{3})$ . The assumption isn't necessary, though it slightly simplifies the work.

\log x y + \log y x = \frac {17}{4} \log x y + \log y x = \frac {16}{4} + \frac {1} {4} Thus, x^4 = y or y^4 = x

since x and y are interchangeable, choose x^4 = y

xy = 288\sqrt{3} x \times x^4 = 288\sqrt{3} x^5 = 288\sqrt{3} x = \sqrt{12}

y= x^4 y = (\sqrt{12})^4 y= 144

x+y = \sqrt{12} + 144 x+y = 144 + \sqrt{12} x+y = 144 + 2\sqrt{3}

a= 144 b= 2 c = 3

a+b+c = 144 + 2 + 3 a+b+c = 149

This solution is not standard. But this is what I did: Given xy=288root3=a+brootc ==>c=3 288=72x4=144x2 Let a=72 b=4 sum=79 Let a=144 b=2 sum=149 One of these two must be answer. Brilliant gives you three attempts. You get answer in two attempts ; that too by calculating within two seconds. 😉

If we take Log (y) to the base x=a,then 4a^2-17a+4=0ie a=4 or 1/4.Sub a=4 in logy to the basex=4 We get y=144 and x=2×root3.x+y=144+2×root3 .then a=144,b=2,c=3.so a+b+c=149.solved☺you dont take a=1/4 because xy=288×root3.