Error or no error

Algebra Level 2

( x 37 ) a + 37 = ( x + a a + 37 ) a \left(\frac x{37}\right)^{a+37}= \left(\frac {x+a}{a+37}\right)^a

Given that a a is a positive integer, find the real x x satisfying the equation above.


The answer is 37.

A Former Brilliant Member by A Former Brilliant Member

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Max Patrick
Dec 4, 2019

Horribly complicated binomial expansions!

I wondered if it is possible to make the Bases (the numbers without the exponent/power/indices ) equal to 1 and then the exponents won't matter.

By setting x=37 for the LHS, then a=0 for the RHS, it is!

0 Helpful 0 Interesting 0 Brilliant 0 Confused

No need to set a = 0 a=0 , it is mentioned that a a is greater than 0 0 . The value x = 37 x=37 satisfies the equation for all positive integers a a

A Former Brilliant Member - 1 year, 6 months ago

Indeed! I've overcomplicated it!

Max Patrick - 1 year, 6 months ago

Just by inspection. Putting x = 37 x=37 , we have the LHS ( x 37 ) a + 37 = 1 a + 37 = 1 \left(\dfrac x{37}\right)^{a+37} = 1^{a+37}=1 and the RHS ( x + a a + 37 ) a = 1 a = 1 \left(\dfrac {x+a}{a+37}\right)^a = 1^a = 1 for all real a a , hence LHS = = RHS, when x = 37 x = \boxed{37} .

0 Helpful 0 Interesting 0 Brilliant 1 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...