Given that 2 log base2 (x+15) - log base2 x = 6
1.Show that x^2 -34x +225 = 0
2. Hence or otherwise solve the equation 2 log base 2 (x+15) - log base 2 x =6
1. To show that x^2 - 34x + 225 = 0, we'll first simplify the equation 2 log base2 (x+15) - log base2 x = 6.
Using the properties of logarithms, we can rewrite the equation as a single logarithmic expression:
log base2 ((x+15) ^2 / x) = 6
We can then rewrite this equation using the definition of logarithms:
2^(6) = ((x+15) ^2 / x)
Simplifying further, we have:
64 = (x+15) ^2 / x
Multiplying both sides by x, we get:
64x = (x+15) ^2
Expanding the square on the right side, we have:
64x = x^2 + 30x + 225
Rearranging terms, we get:
x^2 - 34x + 225 = 0
So, we have shown that x^2 - 34x + 225 = 0.
2. To solve the equation 2 log base2 (x+15) - log base2 x = 6, we can use the fact that log base a (b) = c is equivalent to a^c = b.
Using this, we can rewrite the equation as:
2^6 = (x+15) ^2 / x
Simplifying, we have:
64 = (x+15) ^2 / x
Multiplying both sides by x, we get:
64x = (x+15) ^2
Taking the square root of both sides, we have:
8√x = x+15
Squaring both sides once again, we get:
64x = x^2 + 30x + 225
Rearranging terms, we have:
x^2 - 34x + 225 = 0
We have already shown in part 1 that this equation factors to:
(x-9)(x-25) = 0
Therefore, the solutions for x are:
x = 9 or x = 25.
To show that x^2 - 34x + 225 = 0, we'll start by rewriting the equation given:
2 log base2 (x+15) - log base2 x = 6
Using the logarithmic properties, we can rewrite this equation as:
log base2 ((x+15)^2) - log base2 x = 6
Next, using the quotient rule of logarithms, we can rewrite the equation as a single logarithm:
log base2 ((x+15)^2 / x) = 6
Now, we can convert the equation into exponential form:
2^6 = (x+15)^2 / x
Simplifying, we have:
64 = (x+15)^2 / x
Cross-multiplying, we get:
64x = (x+15)^2
Expanding the right side of the equation, we have:
64x = x^2 + 30x + 225
Rearranging the equation to bring all terms to one side, we get:
x^2 + 30x - 64x + 225 - 0 = 0
Simplifying further, we have:
x^2 - 34x + 225 = 0
Thus, we have shown that x^2 - 34x + 225 = 0.
Moving on to question 2, we'll solve the equation 2 log base2 (x+15) - log base2 x = 6 using the quadratic equation:
x^2 - 34x + 225 = 0
To solve this equation, we can use factoring or the quadratic formula. In this case, the equation can be factored as:
(x - 9)(x - 25) = 0
Setting each factor equal to zero, we have:
x - 9 = 0 or x - 25 = 0
Solving for x, we get:
x = 9 or x = 25
So the solutions to the equation are x = 9 and x = 25.
To prove that x^2 -34x + 225 = 0, let's start by solving the equation 2log(base2)(x+15) - log(base2)x = 6.
Step 1: Combine the logarithms using the properties of logarithms.
Using the log(property3) property of logarithms: a*log(baseM)(N) = log(baseM)(N^a)
2log(base2)(x+15) - log(base2)x = 6
log(base2)((x+15)^2) - log(base2)x = 6
Using the log(property2) property of logarithms: log(baseM)(N) - log(baseM)(P) = log(baseM)(N/P)
log(base2)(((x+15)^2) / x) = 6
Step 2: Apply exponential function to both sides.
Using the definition of logarithms, log(baseM)(N) = P is equivalent to saying that M^P = N.
(base2)^6 = ((x+15)^2) / x
64 = ((x+15)^2) / x
Multiply both sides by x to clear the fraction:
64x = (x+15)^2
Step 3: Expand the equation.
64x = (x+15)(x+15)
64x = x^2 + 30x + 225
Step 4: Rearrange the equation to make it quadratic.
0 = x^2 + 30x + 225 - 64x
0 = x^2 -34x + 225
Therefore, we have shown that x^2 -34x + 225 = 0.
To solve the equation x^2 -34x + 225 = 0:
We can factor the quadratic equation to find the values of x. However, this quadratic equation is not easily factorable.
Alternatively, we can use the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / (2a)
For the equation x^2 -34x + 225 = 0,
a = 1, b = -34, and c = 225.
x = [-(-34) ± sqrt((-34)^2 - 4(1)(225))] / (2*1)
x = [34 ± sqrt(1156 - 900)] / 2
x = [34 ± sqrt(256)] / 2
x = [34 ± 16] / 2
Therefore, the solutions to the equation x^2 -34x + 225 = 0 are:
x1 = (34 + 16) / 2 = 25
x2 = (34 - 16) / 2 = 9
Hence, the solution to the equation 2log(base2)(x+15) - log(base2)x = 6 is x = 9 or x = 25.
2 log base2 (x+15) - log base2 x = 6
log(x+15)^2 - logx = 6
(x+15)^2/x = 2^6
(x+15)^2 = 64x
x^2+30x+225 = 64x
x^2-34x+225 = 0