Logarithm intersection

Algebra Level 2

Find the crossing points of y = log 2 ( x ) y=\log_2(x) and 2 x + y 5 = 0 2x+y-5=0

Give your answer in the form of x + y x+y


The answer is 3.

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2 solutions

The two equations combined yields x × 4 x = 32 x\times 4^x=32 . Since for negative x x the value of the L. H. S. is negative, the curve y = x × 4 x y=x\times 4^x can not intersect the line y = 32 y=32 . For positive x x , the function y = x × 4 x y=x\times 4^x is increasing and at x = 0 x=0 , y = 0 < 32 y=0<32 . Hence it will intersect the line y = 32 y=32 only once in this domain. This will happen when x = 2 x=2 , and hence y = 5 2 x = 1 y=5-2x=1 .

So x + y = 2 + 1 = 3 x+y=2+1=\boxed 3 .

Aryan Sanghi
Jul 2, 2020

By brute force

At x = 2 , y = 1 in both the equations. \text{At } x = 2, y = 1 \text{ in both the equations.}

So, x + y = 3 \text{So, }\color{#3D99F6}{\boxed{x + y = 3}}

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