3^x=9^x+5

Answer 1

To solve the equation \(3^x = 9^x + 5\), we start by expressing \(9^x\) in terms of base \(3\):

\[ 9^x = (3^2)^x = (3^x)^2 \]

So, we can rewrite the equation as:

\[ 3^x = (3^x)^2 + 5 \]

Now, let's introduce a substitution to simplify our equation. Let \(y = 3^x\). Then we can rewrite the equation as:

\[ y = y^2 + 5 \]

Rearranging gives us a quadratic equation:

\[ y^2 - y + 5 = 0 \]

Next, we can apply the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In our case, \(a = 1\), \(b = -1\), and \(c = 5\):

\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - 20}}{2} \]

\[ y = \frac{1 \pm \sqrt{-19}}{2} \]

Since \(\sqrt{-19}\) is imaginary, this implies:

\[ y = \frac{1 \pm i\sqrt{19}}{2} \]

This means \(y\) (or \(3^x\)) takes complex values. Because \(3^x\) is always positive for real \(x\), there are no real solutions to the original equation \(3^x = 9^x + 5\).

Thus, the equation has no real solutions.

Answer 2

Whats the answer for x

Answer 3

Given the earlier steps we derived from the equation \(3^x = 9^x + 5\), we converted it to the quadratic form:

\[ y^2 - y + 5 = 0 \]

Substituting \(y = 3^x\) results in the quadratic equation:

\[ y^2 - y + 5 = 0 \]

Using the quadratic formula:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - 20}}{2} = \frac{1 \pm \sqrt{-19}}{2} \]

This yields:

\[ y = \frac{1 \pm i\sqrt{19}}{2} \]

Since \(y = 3^x\) must be a positive real number for real \(x\), and our results for \(y\) are complex (specifically, they have imaginary parts), we conclude that there are no real solutions for \(x\).

Therefore, the answer is:

There are no real solutions for \(x\).

Answer 4

10
5
-10
-5

Answer 5

It seems like you may be providing potential answer values or choices. Let's check if any of the supplied values: \(10\), \(5\), \(-10\), or \(-5\), satisfy the original equation \(3^x = 9^x + 5\).

1. For \(x = 10\): \[ 3^{10} = 59049, \quad 9^{10} = (3^2)^{10} = 59049 \Rightarrow 9^{10} + 5 = 59049 + 5 = 59054 \] Not equal.

2. For \(x = 5\): \[ 3^{5} = 243, \quad 9^{5} = (3^2)^{5} = 243 \Rightarrow 9^{5} + 5 = 243 + 5 = 248 \] Not equal.

3. For \(x = -10\): \[ 3^{-10} = \frac{1}{59049} \approx 0.00001693, \quad 9^{-10} = \frac{1}{59049} \Rightarrow 9^{-10} + 5 \approx 0.00001693 + 5 \approx 5.00001693 \] Not equal.

4. For \(x = -5\): \[ 3^{-5} = \frac{1}{243} \approx 0.004115225, \quad 9^{-5} = \frac{1}{243} \Rightarrow 9^{-5} + 5 \approx 0.004115225 + 5 \approx 5.004115225 \] Not equal.

None of the values \(10\), \(5\), \(-10\), or \(-5\) satisfy the original equation \(3^x = 9^x + 5\).

If you are looking for a specific type of solution among these, please clarify! Otherwise, the conclusion remains that there are no real solutions for \(x\).

Answer 6

+5 at the end is part of the equation

Answer 7

Thank you for the clarification! Let's solve the equation again, taking into account that \(+5\) is part of the equation.

The original equation is:

\[ 3^x = 9^x + 5 \]

First, we rewrite \(9^x\) as \((3^2)^x = (3^x)^2\):

\[ 3^x = (3^x)^2 + 5 \]

Let \(y = 3^x\). Substituting gives:

\[ y = y^2 + 5 \]

Rearranging this equation, we get:

\[ y^2 - y + 5 = 0 \]

Next, we will use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = -1\), and \(c = 5\):

\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - 20}}{2} = \frac{1 \pm \sqrt{-19}}{2} \]

As calculated previously, this results in:

\[ y = \frac{1 \pm i\sqrt{19}}{2} \]

Since \(y = 3^x\) must be a real and positive value, and we have complex solutions, it confirms there are no real solutions to the original equation \(3^x = 9^x + 5\).

Now let’s evaluate the specific practice values you provided. I'll check them to see if they yield a solution for the original equation:

1. For \(x = 10\): \[ 3^{10} = 59049, \quad 9^{10} = 59049 \Rightarrow 9^{10} + 5 = 59049 + 5 = 59054 \quad (3^{10} \neq 9^{10} + 5) \]

2. For \(x = 5\): \[ 3^{5} = 243, \quad 9^{5} = 59049 \Rightarrow 9^{5} + 5 = 243 + 5 = 248 \quad (3^{5} \neq 9^{5} + 5) \]

3. For \(x = -10\): \[ 3^{-10} = \frac{1}{59049}, \quad 9^{-10} = \frac{1}{59049} \Rightarrow 9^{-10} + 5 \approx 0 + 5 = 5 \quad (3^{-10} \neq 9^{-10} + 5) \]

4. For \(x = -5\): \[ 3^{-5} = \frac{1}{243}, \quad 9^{-5} = \frac{1}{243} \Rightarrow 9^{-5} + 5 \approx 0 + 5 = 5 \quad (3^{-5} \neq 9^{-5} + 5) \]

After checking each proposed value, none of them satisfy the given equation.

Therefore: None of the provided values for \(x\) (10, 5, -10, -5) satisfy the equation \(3^x = 9^x + 5\).

If you have a different equation in mind or if you need more assistance, please let me know!

Answer 8

WHY ARE YOU SO GODDANG DUMB
3^x=9^x+5 that plus is up with the x

Answer 9

I apologize for the misunderstanding earlier. Thank you for your patience and for clarifying the structure of the equation. The correct interpretation of the equation is:

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

Now, let's solve this correctly. First, recall that \(9\) can be rewritten as \(3^2\). Thus:

\[ 9^{x+5} = (3^2)^{x+5} = 3^{2(x+5)} = 3^{2x + 10} \]

Now, we can rewrite the original equation:

\[ 3^x = 3^{2x + 10} \]

Since the bases are the same, we can set the exponents equal to each other:

\[ x = 2x + 10 \]

To isolate \(x\), we'll rearrange the equation:

\[ x - 2x = 10 \implies -x = 10 \implies x = -10 \]

So, the solution to the equation \(3^x = 9^{x+5}\) is:

\[ \boxed{-10} \]