We don’t like the log_5, so raise both sides as power of 5, remembering your Indices Laws…

(1-2x) = 5 / (2-x)

The rest is easy, you can see it’s a quadratic:

(1-2x)(2-x) = 5

2x² -5x +2 = 5

2x² -5x -3 = 0

(2x+1)(x-3) = 0 [got this wrong first time]

x= [5 ± √25-4(2)(-3)] / 2(2)

= [5 ± √49] / 2(2)

= (5-7) /4 or (5+7)/4

x = -1/2 or 3 [you might be able]

* How do you solve this: sin(2x) = -cos(x) in [0,2Pi)

Knowing your identity for sin(2x)…

2sin(x)cos(x) = -cos(x)

(2sin(x) + 1) cos(x) = 0

[Common mistake: do not ignore the cos(x) = 0 solutions]

cos(x) = 0 => solutions are x=0, pi

sin(x) = -1/2 => solutions are in 3rd or 4th quadrant

x=(pi +pi/6)=7pi/6, (2pi – pi/6)=11pi/6

Solutions are x=0, pi, 7pi/6, 11pi/6