To solve this logarithmic equation, we can use the properties of logarithms.
The equation is:log4(5x – 4) + log4(5x – 1) = 1
We can combine the two logarithms into a single logarithm by using the product rule of logarithms which states that log(a) + log(b) = log(ab).
log4((5x – 4)(5x – 1)) = 1log4(25x^2 - 9x - 20) = 1
To eliminate the logarithm, we need to rewrite the equation in exponential form. Remember that loga(b) = c is equivalent to a^c = b.
So, using this in our equation:
4^1 = 25x^2 - 9x - 204 = 25x^2 - 9x - 2025x^2 - 9x - 24 = 0
Now, we need to solve the quadratic equation 25x^2 - 9x - 24 = 0. This equation can be factored as:
(5x + 4)(5x - 6) = 0
Setting each factor to zero:
5x + 4 = 05x = -4x = -4/5
and
5x - 6 = 05x = 6x = 6/5
So, the solutions to the equation log4(5x – 4) + log4(5x – 1) = 1 are x = -4/5 and x = 6/5.
To solve this logarithmic equation, we can use the properties of logarithms.
The equation is:
log4(5x – 4) + log4(5x – 1) = 1
We can combine the two logarithms into a single logarithm by using the product rule of logarithms which states that log(a) + log(b) = log(ab).
log4((5x – 4)(5x – 1)) = 1
log4(25x^2 - 9x - 20) = 1
To eliminate the logarithm, we need to rewrite the equation in exponential form. Remember that loga(b) = c is equivalent to a^c = b.
So, using this in our equation:
4^1 = 25x^2 - 9x - 20
4 = 25x^2 - 9x - 20
25x^2 - 9x - 24 = 0
Now, we need to solve the quadratic equation 25x^2 - 9x - 24 = 0. This equation can be factored as:
(5x + 4)(5x - 6) = 0
Setting each factor to zero:
5x + 4 = 0
5x = -4
x = -4/5
and
5x - 6 = 0
5x = 6
x = 6/5
So, the solutions to the equation log4(5x – 4) + log4(5x – 1) = 1 are x = -4/5 and x = 6/5.