Use logarithms to solve the given equation. (Round the answer to four decimal places.)

9(1.055*x* + 1) = 10

*x* =

We are given the equation:

\[
9(1.055^x + 1) = 10
\]

### Step 1: Isolate the exponential expression
First, we divide both sides of the equation by 9:

\[
1.055^x + 1 = \frac{10}{9}
\]

\[
1.055^x + 1 = 1.1111
\]

Now subtract 1 from both sides:

\[
1.055^x = 0.1111
\]

### Step 2: Apply logarithms
To solve for \(x\), take the logarithm of both sides. We will use the natural logarithm (ln), though you can also use the logarithm base 10 (log):

\[
\ln(1.055^x) = \ln(0.1111)
\]

Using the logarithmic property \(\ln(a^b) = b \cdot \ln(a)\), we rewrite the left-hand side:

\[
x \cdot \ln(1.055) = \ln(0.1111)
\]

### Step 3: Solve for \(x\)
Now, solve for \(x\) by dividing both sides by \(\ln(1.055)\):

\[
x = \frac{\ln(0.1111)}{\ln(1.055)}
\]

### Step 4: Calculate the result
Let me calculate this value for you.The value of \(x\), rounded to four decimal places, is approximately \(-41.0402\).

Would you like more details on this process or have any questions?

Here are 5 related questions you could explore:
1. How do logarithmic properties help in solving exponential equations?
2. Can logarithms with different bases (e.g., base 10 vs. natural log) lead to the same solution?
3. What is the significance of the base in the exponential equation?
4. How does the value of \(x\) change if the base 1.055 were larger or smaller?
5. How can logarithms be used to simplify more complex exponential equations?

**Tip:** When solving exponential equations, using logarithms allows us to "bring down" the exponent, making the equation linear and easier to solve.

Use logarithms to solve the given equation. (Round the answer to four decimal places.)

9(1.055^x + 1) = 10

x = ?

Learn how to solve the exponential equation 9(1.055^x + 1) = 10 using logarithms. This step-by-step guide covers key algebraic concepts and the logarithmic theorem, with a detailed explanation and calculations rounded to four decimal places. Ideal for students in Grades 10-12.

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