Insane Exponential Equation

[nycmathguy]

Solve for x.

(6 - sqrt{35})^(x) + (6 + sqrt{35})^(x) = 142

 

[MathLover1]

(6 - sqrt(35))^x+ (6 + sqrt(35))^x = 142

apply exponent rules

((6 + sqrt(35))^x)^-1+ (6 + sqrt(35))^x = 142

rewrite the equation with (6 + sqrt(35))^x=u

u^-1+u=142...........solve for u

1/u+u=142
(1+u^2)/u=142
1+u^2=142u
u^2-142u+1=0......using quadratic formula we get

u = 71+ 12 sqrt(35)
u = 71 - 12 sqrt(35)

substitute back u=(6+ sqrt(35))^x, solve for x

(6+ sqrt(35))^x = 71+ 12 sqrt(35)

log((6+ sqrt(35))^x )=log( 71+ 12 sqrt(35))

x*log((6+ sqrt(35) )=log( 71+ 12 sqrt(35))

x=log( 71+ 12 sqrt(35))/log((6+ sqrt(35) )

x=2

(6+ sqrt(35))^x = 71- 12 sqrt(35)

log((6+ sqrt(35))^x )=log( 71- 12 sqrt(35))

x*log((6+ sqrt(35) )=log( 71-12 sqrt(35))

x=log( 71- 12 sqrt(35))/log((6+ sqrt(35) )

x=-2

 

[nycmathguy]

(6 - sqrt(35))^x+ (6 + sqrt(35))^x = 142

apply exponent rules

((6 + sqrt(35))^x)^-1+ (6 + sqrt(35))^x = 142

rewrite the equation with (6 + sqrt(35))^x=u

u^-1+u=142...........solve for u

1/u+u=142
(1+u^2)/u=142
1+u^2=142u
u^2-142u+1=0......using quadratic formula we get

u = 71+ 12 sqrt(35)
u = 71 - 12 sqrt(35)

substitute back u=(6+ sqrt(35))^x, solve for x

(6+ sqrt(35))^x = 71+ 12 sqrt(35)

log((6+ sqrt(35))^x )=log( 71+ 12 sqrt(35))

x*log((6+ sqrt(35) )=log( 71+ 12 sqrt(35))

x=log( 71+ 12 sqrt(35))/log((6+ sqrt(35) )

x=2

(6+ sqrt(35))^x = 71- 12 sqrt(35)

log((6+ sqrt(35))^x )=log( 71- 12 sqrt(35))

x*log((6+ sqrt(35) )=log( 71-12 sqrt(35))

x=log( 71- 12 sqrt(35))/log((6+ sqrt(35) )

x=-2

Very impressive math work.