I need help, can someone guide me through this problem?
log3(x + 2) + log3(4x – 1) = 5
sum of logs is log of product
log3 (4 x^2 + 7 x -2) = 5
base ^logbase(z) = z
so
3^[log3 (4 x^2 + 7 x -2)] = 3^5
or
(4 x^2 + 7 x -2) = 243
4 x^2 + 7 x - 245 = 0
(x-7)( 4x+35) = 0
x = 7 or x = -35/4
thank u so much!!! honestly u don't know how much u helped me :)
You are welcome, good luck :)
Sure! I can guide you through the problem.
To solve the equation log3(x + 2) + log3(4x – 1) = 5, we can apply logarithm properties to simplify the equation. The logarithm property we'll use is the product rule, which states that loga(x) + loga(y) = loga(xy).
Using the product rule, we can rewrite the equation as a single logarithm: log3((x + 2)(4x – 1)) = 5.
Next, we'll use another logarithm property called the exponent rule, which states that loga(x) = y is equivalent to ax = x.
Applying the exponent rule, we get 3^5 = (x + 2)(4x – 1).
Simplifying further, we have 243 = (x + 2)(4x – 1).
To solve for x, we'll expand the right side of the equation: 243 = 4x^2 + 7x - 2.
Setting the equation equal to zero, we have 4x^2 + 7x - 245 = 0.
Now, we can use factoring, completing the square, or the quadratic formula to solve for x.
Factoring the equation, we have (x + 15)(4x - 16) = 0.
Setting each factor equal to zero, we have two possible solutions:
x + 15 = 0 and 4x - 16 = 0.
Solving each equation, we get x = -15 and x = 4.
Therefore, the solutions to the equation log3(x + 2) + log3(4x – 1) = 5 are x = -15 and x = 4.