Determine the exact value of x.
a. 3^2x = 5(3^x) +36
b. (1/8)^x-3 = 2x16^2x+1
c. 3^x2 + 20 = (1/27)^3X
a) let 3^x = y
so you have y^2 - 5y - 36=0
(y-9)(y+4) = 0
y = 9 or y = -4
3^x = 9
3^x = 3^2
x = 2
or 3^x = -4 , not possible
b) the trick here is to see that all bases are powers of 2
1/8 = 2^-3
16 = 2^4
so (1/8)^(x-3) = 2x16^(2x+1)
(2^-3)^(x-3) = 2(2^4)^(2x+1)
2^(-3x+9) = 2(2)^(8x+4)
2^(-3x+9) = 2^(8x+5)
then
-3x+9 = 8x+5
you can finish it ...
c) I see no easy way to do this one.
Why is the a capital X ?
Thank you for the help,
I'm really stuck on c. The x was made capital by mistake, it's actually supposed to be lower case.
The equation is actually 3^x^2 + 20 = (1/27)^3x
To determine the exact value of x in each of the given equations, we'll need to simplify the equations and solve for x. Let's go through each equation one by one:
a. 3^2x = 5(3^x) + 36:
To simplify this equation, let's divide both sides of the equation by 3^x:
(3^2x) / (3^x) = (5 * 3^x + 36) / (3^x)
Using the exponent rule for division (subtracting exponents with the same base), we have:
3^(2x - x) = (5 * 3^x + 36) / (3^x)
Simplifying further:
3^x = (5 * 3^x + 36) / (3^x)
To isolate the variable x, let's multiply both sides of the equation by 3^x:
3^x * 3^x = 5 * 3^x + 36
Using the exponent rule for multiplication (adding exponents with the same base), we have:
3^(2x) = 5 * 3^x + 36
Now, this equation does not have an obvious algebraic solution. It may require numerical methods or approximations to find the exact value of x.
b. (1/8)^(x-3) = 2x * 16^(2x+1):
To simplify this equation, let's deal with each side separately.
On the left side, we have a fraction raised to an exponent. We can rewrite 1/8 as (2^-3) using exponent rules, then use the rule for exponentiation of fractions:
(2^-3)^(x-3) = 2x * 16^(2x+1)
Applying the exponent rule for exponentiation of exponents, we get:
2^(-3 * (x-3)) = 2x * 16^(2x+1)
Simplifying the left side:
2^(-3x + 9) = 2x * 16^(2x+1)
Now, let's deal with the right side of the equation. Recall that 16 is equal to 2^4. Substituting this in:
2^(-3x + 9) = 2x * (2^4)^(2x+1)
Using the exponent rule for exponentiation (multiplying exponents when raising an exponent to another exponent), we have:
2^(-3x + 9) = 2x * 2^(4 * (2x+1))
Applying the exponent rule for multiplication of exponents, we get:
2^(-3x + 9) = 2x * 2^(8x + 4)
Now, we have both sides of the equation written with the same base (2). To equate the exponents, we can drop the base and equate the exponents:
-3x + 9 = 8x + 4
To solve for x, we can simplify and solve the resulting linear equation. Subtracting 8x from both sides and adding 3x to both sides:
9 - 4 = 8x + 3x
5 = 11x
Dividing both sides by 11:
x = 5/11
c. 3^(x^2) + 20 = (1/27)^(3x):
To simplify this equation, let's focus on the right side first. Recall that 1/27 is equal to 3^(-3). Substituting this in:
3^(x^2) + 20 = (3^(-3))^(3x)
Applying the exponent rule for exponentiation of exponents, we have:
3^(x^2) + 20 = 3^(-3 * (3x))
Simplifying the right side:
3^(x^2) + 20 = 3^(-9x)
Now, both sides of the equation are written with the same base (3). To equate the exponents, we can drop the base and equate the exponents:
x^2 = -9x
Adding 9x to both sides:
x^2 + 9x = 0
Factoring the left side:
x(x + 9) = 0
To solve for x, we can set each factor equal to zero:
x = 0 or x + 9 = 0
x = 0 or x = -9
Therefore, in equation c, x can take on the values 0 or -9.