Logb(x-1) + logb(x+2) = logb(8-2x)

x=2

Step-by-step explanation:

log_x(y)+log_x(z)=log_x(yz), so logb(x-1) + logb(x+2)=logb((x-1)(x+2)). next subtract logb(8-2x) from both sides to get logb((x-1)(x+2))-logb(8-2x)=0. log_x(y) - log_x(z) = log_x(y/z). so now we have logb((x-1)(x+2)/(8-2x)). now you can put it into exponential form: (x-1)(x+2)/(8-2x)=b^0=1 now just solve for x:

(x-1)(x+2)= 8-2x, x^2 + x -2 = 8-2x, x^2 + 3x -10 = 0, (x+5)(x-2)=0 x=-5, x=2. plug both into the original equation to check which one is correct, since log_x(y) can't have a negative y, x=-5 doesnt work

step-by-step explanation:

for the free