(1) You have to recall some laws of exponents.
15 * 3^(y+1) - 243 * 5^(y-2) = 0
3 * 5 * 3^(y+1) - 3^5 * 5^(y-2) = 0
5 * 3^(y+1+1) - 3^5 * 5^(y-2) = 0
5 * 3^(y+2) = 3^5 * 5^(y-2)
3^(y+2) / 3^5 = 5^(y-2) / 5
3^(y+2-5) = 5^(y-2-1)
3^(y-3) = 5^(y-3)
The only number that will satisfy this is when the exponent is zero. When the exponent of both is zero, they will be equal to 1. thus,
y - 3 = 0
y = 3
(2) Not sure if that y-1 is entirely inside the log, or only y is inside the log. But I solved for the one where only y is inside:
log (y) - 1 = -log (y-9)
log (y) - log (10) = log (1/(y-9))
log (y/10) = log (1/(y-9))
Equating the terms inside the log:
y/10 = 1/(y-9)
Cross-multiply:
y^2 - 9y = 10
y^2 - 9y - 10 = 0
Factoring:
(y-10)(y+1) = 0
y = 10
y = -1
If you substitute back these roots, you'll see that the only root that will satisfy is the positive root (y = 10), because the term inside the log cannot be negative (will happen if you substitute y = -1).
Hope this helps~ `u`
1. Solve for y in the equation 15 * 3^(y+1) - 243 * 5^(y-2) = 0.
2. Solve: log y-1 = -log(y-9)
2 answers
Thank you very much. Very clear explanation.:)
2 answers