exponentiating to solve equations: solve the eaquation, cheack for extraneous solutions
log5(x+4)+log5(x+1)=2
remember log a + log b= log ab
do that, then take the antilog..
log5 (x+4)(x+1)=2
(x+4)(x+1)=25
then proceed to solve.
log5 [(x+4)(x+1)] = 2
5^log5 [(x+4)(x+1)] = 5^2 = 25
(x+4)(x+1) = 25
x^2 + 5 x + 4 = 25
x^2 + 5 x - 21 = 0
x = [-5 +/- sqrt(25+84) ]/2
= [-5 +/- sqrt(109)]/2
= [ -2.5 +/- 5.22 ]
= 2.72 or - 7.72
You can not take log of a negative number so use 2.72
i dnt get bwat u did after x^2+5x-21=0. that is were i keep gettin stuck on
http://en.wikipedia.org/wiki/Quadratic_equation
To solve the equation log5(x+4) + log5(x+1) = 2 and check for extraneous solutions, we can use the properties of logarithms and exponentiation. Here's how you can do it step by step:
Step 1: Combine the logarithms using the product rule of logarithms.
log5((x+4)(x+1)) = 2
Step 2: Rewrite the equation in exponential form.
5^2 = (x+4)(x+1)
Step 3: Simplify and solve the quadratic equation.
25 = x^2 + 5x + 4
x^2 + 5x - 21 = 0
Step 4: Factor the quadratic equation or use the quadratic formula to find the solutions. In this case, factoring won't be easy, so let's use the quadratic formula.
x = (-b ± √(b^2 - 4ac))/(2a)
For our equation x^2 + 5x - 21 = 0, the coefficients are a = 1, b = 5, and c = -21.
x = (-5 ± √(5^2 - 4(1)(-21)))/(2(1))
x = (-5 ± √(25 + 84))/2
x = (-5 ± √109)/2
So we have two possible solutions for x:
x1 = (-5 + √109)/2
x2 = (-5 - √109)/2
Step 5: Check for extraneous solutions.
To check for extraneous solutions, substitute each value of x back into the original equation and make sure the logarithmic expressions are valid. In this case, logarithms are only valid for positive numbers.
For x1 = (-5 + √109)/2:
log5(((-5 + √109)/2) + 4) + log5(((-5 + √109)/2) + 1) = 2
Use a calculator to compute each logarithmic expression:
log5(1.25) + log5(0.75) = 2
0.602 + (-0.12) = 2
0.482 ≠ 2
The equation is not satisfied, so x1 = (-5 + √109)/2 is not a solution.
For x2 = (-5 - √109)/2:
log5(((-5 - √109)/2) + 4) + log5(((-5 - √109)/2) + 1) = 2
Using a calculator again:
log5(7.25) + log5(2.25) = 2
0.860 + 0.431 = 2
1.291 = 2
The equation is still not satisfied, so x2 = (-5 - √109)/2 is also not a solution.
Therefore, the original equation log5(x+4) + log5(x+1) = 2 has no real solutions.