Solve for x in the following equations. (4 marks)
log3(x - 5) + log3(x + 3) = 2 <--- both log3.. 3 is subscript
log7x + log7(x - 1) = log72x <--- all log7.. 7 is subscript.
use your log rules:
log (AB) = logA + logB
and
log (A/B) = logA - logB
1st one:
log3 [(x-5)(x+3)] = 2
by definition of logs
(x-5)(x+3) = 3^2 = 9
x^2 - 2x -24=0
(x-6)(x+4) = 0
x = 6 or x=-4 , but x=-4 would make log3(x+3) undefined, so
x = 6
2nd:
log7(x) + log7(x-1) - log7(2x) = 0
log7[x(x-1)/(2x)] = 0
log7[(x-1)/2] = 0
(x-1)/2 = 7^0
(x-1)/2 = 1
x-1=2
x = 3
To solve for x in the given equations, we'll use logarithmic properties and properties of equality.
Equation 1: log3(x - 5) + log3(x + 3) = 2
Step 1: Combine the two logarithms by using the product rule of logarithms, which states that log(a) + log(b) = log(ab).
log3((x - 5)(x + 3)) = 2
Step 2: Rewrite the equation in exponential form. In this case, since the base is 3, we rewrite it as 3^2 = (x - 5)(x + 3).
9 = (x - 5)(x + 3)
Step 3: Expand and rearrange the equation.
9 = x^2 - 2x - 15
x^2 - 2x - 24 = 0
Step 4: Solve the quadratic equation by factoring or using the quadratic formula.
(x - 6)(x + 4) = 0
x - 6 = 0 or x + 4 = 0
x = 6 or x = -4
Therefore, the solutions for x in Equation 1 are x = 6 and x = -4.
Equation 2: log7x + log7(x - 1) = log72x
Step 1: Combine the two logarithms using the sum rule of logarithms, which states that log(a) + log(b) = log(ab).
log7(x(x - 1)) = log72x
Step 2: Rewrite the equation in exponential form. In this case, since the base is 7, we rewrite it as 7^(log72x) = x(x - 1).
7^(log72x) = x(x - 1)
Step 3: Simplify the equation by using the fact that 7^(log72x) = 2x.
2x = x(x - 1)
Step 4: Expand and rearrange the equation.
2x = x^2 - x
x^2 - 3x = 0
Step 5: Solve the quadratic equation by factoring or using the quadratic formula.
x(x - 3) = 0
x = 0 or x - 3 = 0
x = 0 or x = 3
Therefore, the solutions for x in Equation 2 are x = 0 and x = 3.