1) Solve for x in terms of k:
log3(x) + log3(x+7) = k; 3 is the base
2) e^(x+5) = e^(x) + 6, solve for x.
Please help, I have no idea how to do these and I've tried several times!
Sure, I can help you with these equations step by step!
1) Solve for x in terms of k:
First, apply the logarithmic property that states the sum of logarithms is equal to the logarithm of the product.
So, we rewrite the equation as:
log3(x(x+7)) = k
Next, apply the exponentiation property that states if logb(x) = y, then b^y = x.
So, using this property, we can rewrite the equation as:
3^k = x(x+7)
Now, let's solve for x by using the quadratic formula.
The equation is in the form of ax^2 + bx + c = 0, where a = 1, b = 7, and c = -3^k.
Using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
Substituting the values, we have:
x = (-(7) ± sqrt((7)^2 - 4(1)(-3^k))) / (2(1))
x = (-7 ± sqrt(49 + 12(3^k))) / 2
So, x in terms of k is:
x = (-7 ± sqrt(49 + 36(3^k))) / 2
2) Solve for x in e^(x+5) = e^x + 6:
Start by subtracting e^x from both sides of the equation:
e^(x+5) - e^x = 6
Now, factor out e^x in the left-hand side of the equation:
e^x * e^5 - e^x = 6
Next, combine the terms on the left-hand side:
e^x (e^5 - 1) = 6
To solve for x, divide both sides of the equation by (e^5 - 1):
e^x = 6 / (e^5 - 1)
Finally, take the natural logarithm of both sides to solve for x:
ln(e^x) = ln(6 / (e^5 - 1))
x = ln(6 / (e^5 - 1))
So the solution for x is:
x = ln(6 / (e^5 - 1))
Sure! Let's solve each of the equations step by step.
1) Solve for x in terms of k:
To solve this equation, we can use the properties of logarithms. The given equation is:
log3(x) + log3(x + 7) = k
Using the logarithmic property, we can combine the sum of two logarithms with the same base into a single logarithm:
log3(x(x + 7)) = k
Now, we can rewrite the equation using exponential form:
3^k = x(x + 7)
Simplifying further:
3^k = x^2 + 7x
To solve this quadratic equation for x, we'll set it equal to zero and use the quadratic formula:
x^2 + 7x - 3^k = 0
The quadratic formula is:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 1, b = 7, and c = -3^k. Substituting these values into the quadratic formula:
x = (-7 ± √(7^2 - 4(1)(-3^k))) / (2(1))
Simplifying further will give you the solution for x in terms of k.
2) e^(x+5) = e^x + 6:
To solve this equation, we can start by subtracting e^x from both sides:
e^(x+5) - e^x = 6
Next, we can factor out e^x from the left side of the equation:
e^x (e^5 - 1) = 6
Now, we divide both sides of the equation by (e^5 - 1):
e^x = 6 / (e^5 - 1)
To isolate x, we can take the natural logarithm (ln) of both sides:
ln(e^x) = ln(6 / (e^5 - 1))
Using the logarithmic property, the left side simplifies to x:
x = ln(6 / (e^5 - 1))
Using a calculator, you can find the numerical value for x.
log3x +log3(x+7)=log3(x)(x+7)
then take the antilog of each side.
x(x+7)=3^k then, this will be a quadratic equation.