In this case Melody, it might be better to let y = √x (a positive number) when you get to 0.08x - 1 = √x so that you have 0.08y2 - 1 = y. Then when you solve for y, if you get a negative number you can discard it immediately:
0.08×y2−1=y⇒{y=−(5×√33−25)4y=(5×√33+25)4}⇒{y=−0.9307033081725358y=13.4307033081725358}
Taking only the positive value we then have (approximately)
x=13.43072⇒x=180.38370249
.
log(8x)−log(1+x1/2)=2log8x1+x0.5=210log8x1+x0.5=1028x1+x0.5=1008x=100(1+x0.5)0.08x=1+√x0.08x−1=√x0.0064x2−0.16x+1=x0.0064x2−1.16x+1=0
x is approx 0.866 or 180.384 but these must be checked. :)
0.0064×x2−1.16×x+1=0⇒{x=−(125×√33−725)8x=(125×√33+725)8}⇒{x=0.8662086478433022x=180.3837913521566978}
log(8x)-log(1+x^1/2)=2
check
log10(8×0.866208643)−log10(1+0.8662086430.5)=0.5549969596406122
log10(8×180.383791352)−log10(1+180.3837913520.5)=1.9999999999997984
Sox≈180.184
.In this case Melody, it might be better to let y = √x (a positive number) when you get to 0.08x - 1 = √x so that you have 0.08y2 - 1 = y. Then when you solve for y, if you get a negative number you can discard it immediately:
0.08×y2−1=y⇒{y=−(5×√33−25)4y=(5×√33+25)4}⇒{y=−0.9307033081725358y=13.4307033081725358}
Taking only the positive value we then have (approximately)
x=13.43072⇒x=180.38370249
.