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Algebra Level 4

Let $a,b,c,d$ be positive integers and $\log_a{b}=\frac{3}{2}, \log_c{d}=\frac{5}{4}$ . If $a-c=9$ , what is the value of $b-d$ ?

2 solutions

Christopher Boo
May 5, 2014

From the problem, we know that

$\displaystyle a^\frac{3}{2}=b$

$\displaystyle c^\frac{5}{4}=d$

Given that $a$ and $b$ is a positive integer,

$a$ can be expressed as $x^2$ where $x$ is an integer.

$c$ can be expressed as $y^4$ where $y$ is an integer.

$x^2-y^4=9$

$(x-y^2)(x+y^2)=9$

Clearly $(x+y^2)>(x-y^2)$ , so we cannot have $(x-y^2)=(x+y^2)=3$ , let

$x-y^2=1$

$x+y^2=9$

Solve the system of equation we get

$x=5, y=2$

Substitute the value of $x,y$ to get $b,d$ , we have

$5^3-2^5=\boxed{93}$

same method

Mardokay Mosazghi - 7 years, 1 month ago

very strange way

suraj Upadhyay - 7 years, 1 month ago

good solution

Albertus Djauhari - 7 years, 1 month ago

good solution

Harsh Deswal - 7 years, 1 month ago

was a bouncer... !!!

Rishabh Tripathi - 7 years ago

Your striking power should be good for this

Vipul Kumar - 7 years, 1 month ago

Laith Hameed
Jul 31, 2014

(a^3/2) = b

(c^5/4) = d

c - d = 9 = 25 - 16

b - d = (a^3/2) - (c^5/4) = (25^3/2) - (16^5/4) = 93