Solve each logarithmic equation. Check for extraneous solutions
log42 = log21 +log(x-1)
Just use your log rules:
log42 = log21 +log(x-1)
log 42 = log (21(x-1))
42 = 21(x-1)
etc, make sure x>1 to have a valid answer
Log42 = Log21 + Log(x-1)
Log(x-1) = Log(42)-Log(21) = 0.30103.
x-1 = 10^(0.30103) = 2.0,
X = 2 + 1 = 3.
To solve the logarithmic equation log 42 = log 21 + log(x - 1), we can use the properties of logarithms.
First, let's simplify the equation using the property log a + log b = log (a * b):
log 42 = log (21 * (x - 1))
Next, we can use the property log a = log b to eliminate the logarithms:
42 = 21 * (x - 1)
Now, let's solve for x:
42 = 21x - 21
Adding 21 to both sides:
63 = 21x
Dividing both sides by 21:
3 = x
So, the solution to the logarithmic equation is x = 3.
To check for extraneous solutions, we need to substitute the value of x back into the original equation and verify if it satisfies the equation:
log 42 = log 21 + log(3 - 1)
log 42 = log 21 + log 2
Logarithmic equations have only one solution, therefore, there are no extraneous solutions to check in this case.
Thus, the solution to the equation is x = 3.