Question

Solve the exponential equation 3^2x = 27(3)^-5x-9

Answer 1

To solve the exponential equation 32^x = 27^(3-5x-9), we can simplify it by first expressing both sides with the same base. Let's rewrite 27 as 3^3:

32^x = (3^3)^(3-5x-9)

Now, apply the exponent rule that states (a^b)^c = a^(b*c):

32^x = 3^(3(3-5x-9))

Now, simplify the exponent on the right side:

32^x = 3^(3(3-5x-9)) = 3^(3(3-5x-9)) = 3^(3(3-5x-9))

Now, we can equate the exponents:

x = 3(3-5x-9)

Let's solve for x:

x = 3(3-5x-9)

x = 3(3-5x-9)

x = 3(3-5x-9)

Now, distribute the 3 on the right side:

x = 9 - 15x - 27

Now, combine like terms:

16x = 9 - 27

16x = -18

Now, divide by 16:

x = -18 / 16

Simplify the fraction:

x = -9/8

So, the solution to the exponential equation 32^x = 27^(3-5x-9) is x = -9/8.

Answer 2

To solve the exponential equation 32x = 27(3)-5x-9, we first need to simplify the exponential expression on the right side.

27(3) can be simplified as 3^3, which equals 27.

So, the equation becomes:

32x = 27 - 5x - 9

Combining like terms, we have:

37x = 18

Finally, we can solve for x by dividing both sides of the equation by 37:

x = 18/37

Answer 3

To solve the exponential equation 32^x = 27^(3-5x-9), you can start by simplifying the equation:

32^x = 27^(-5x-6)

Now, you can rewrite 32 as a power of 3 because 27 is also a power of 3:

(3^5)^x = (3^3)^(-5x-6)

Now, apply the properties of exponents:

3^(5x) = 3^(-15x-18)

For two exponential expressions with the same base to be equal, their exponents must be equal. So, set the exponents equal to each other:

5x = -15x - 18

Now, solve for x:

5x + 15x = -18

20x = -18

x = -18 / 20

x = -9/10

So, the solution to the exponential equation 32^x = 27^(3-5x-9) is x = -9/10.

Answer 4

32^x = 27^(3-5x-9)

First, simplify the exponent on the right side:

32^x = 27^(-5x-6)

Now, notice that 32 is equal to 2^5 and 27 is equal to 3^3:

(2^5)^x = (3^3)^(-5x-6)

Apply the power rule for exponents (raising a power to another power):

2^(5x) = 3^(-15x-18)

Now, you have an equation with matching bases (2 and 3). Since the bases are different, you'll need to use logarithms to solve for x. You can use either the natural logarithm (ln) or the common logarithm (log base 10). Let's use the natural logarithm:

Take the natural logarithm of both sides:

ln(2^(5x)) = ln(3^(-15x-18))

Now, apply the power rule for logarithms (ln(a^b) = b * ln(a)):

5x * ln(2) = (-15x-18) * ln(3)

Now, isolate the variable x:

5x * ln(2) = -15x * ln(3) - 18 * ln(3)

Now, move the terms with x to one side of the equation:

5x * ln(2) + 15x * ln(3) = -18 * ln(3)

Factor out x:

x * (5 * ln(2) + 15 * ln(3)) = -18 * ln(3)

Now, divide both sides by the coefficient of x:

x = (-18 * ln(3)) / (5 * ln(2) + 15 * ln(3))

Answer 5

There is no real root. There is imaginary roots. So this question has no real answer. There is no need even thow here is imaginary roots x=-18/16 and x=-9/8.so that's it.

Answer 6

Write 32 as 2^5 and 27 as 3^3

2^5 = 3^3 . 3 ^(-5x+9)

2^5x=3^3-5x-9

3^-5x-6

2^5x=3^-5x-6 

5x = -5x -6

Collect like terms 

10x = -6

x = -6/10 

X = -3/5 

Answer 7

To solve the exponential equation \( 3^{2x} = 27(3)^{-5x-9} \), let's first simplify the right side using the properties of exponents.

We know that \( 27 = 3^3 \). So, we can rewrite \( 27(3)^{-5x-9} \) as \( (3^3)(3)^{-5x-9} \). According to the properties of exponents, when multiplying powers with the same base, we add their exponents. Therefore:

\[ (3^3)(3)^{-5x-9} = 3^{3+(-5x-9)} = 3^{-5x} \]

Now, our equation becomes \( 3^{2x} = 3^{-5x} \).

Since the bases are the same, we can equate the exponents:

\[ 2x = -5x \]

Now, we can solve for \( x \):

\[ 2x + 5x = 0 \]

\[ 7x = 0 \]

\[ x = 0 \]

So, the solution to the exponential equation \( 3^{2x} = 27(3)^{-5x-9} \) is \( x = 0 \).