1. Solve for y in the equation 15 * 3^(y+1) - 243 * 5^(y-2) = 0.
2. Solve: log y-1 = -log(y-9)
(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`
Thank you very much. Very clear explanation.:)
1. To solve for y in the equation 15 * 3^(y+1) - 243 * 5^(y-2) = 0, we can follow these steps:
Step 1: Simplify the equation by expanding the exponents:
15 * 3 * 3^y - 243 * 5 * 5^y = 0
Step 2: Simplify further:
45 * 3^y - 1215 * 5^y = 0
Step 3: Manipulate the equation by dividing both sides by 45:
3^y - 27 * 5^y = 0
Step 4: Notice that there is a common factor of 3^y in both terms. Rewrite the equation as:
3^y(1 - 27 * 5^y) = 0
Step 5: Set each factor equal to zero and solve separately:
3^y = 0
1 - 27 * 5^y = 0
Step 6: Solve the first equation:
3^y = 0 is not possible since any number raised to the power of 0 is 1.
Step 7: Solve the second equation:
1 - 27 * 5^y = 0
27 * 5^y = 1
5^y = 1/27
Step 8: Take the logarithm (base 5) of both sides:
log(5, 5^y) = log(5, 1/27)
y * log(5, 5) = log(5, 1/27)
Step 9: Simplify using the property of logarithms:
y * 1 = log(5, 1/27)
y = log(5, 1/27)
Step 10: Evaluate the logarithm using the change of base formula or a calculator:
y ≈ log(1/27) / log(5)
y ≈ -2.77
Therefore, the solution to the equation is y ≈ -2.77.
2. To solve the equation log(y-1) = -log(y-9), follow these steps:
Step 1: Manipulate the equation by bringing the negative logarithm to the other side:
log(y-1) + log(y-9) = 0
Step 2: Combine the two logarithms into a single logarithm using the product rule:
log((y-1)(y-9)) = 0
Step 3: Rewrite the equation in exponential form:
(y-1)(y-9) = 10^0
(y-1)(y-9) = 1
Step 4: Expand and simplify the equation:
y^2 - 10y + 9 = 1
Step 5: Move all terms to one side of the equation:
y^2 - 10y + 8 = 0
Step 6: Factorize the quadratic equation or use the quadratic formula to solve for y:
(y-2)(y-4) = 0
So, y-2 = 0 or y-4 = 0
Step 7: Solve each equation separately:
y-2 = 0
y = 2
and
y-4 = 0
y = 4
Therefore, the solutions to the equation are y = 2 and y = 4.