To find the intercepts of the equation $y = \log(3x + 4) + 2$, we proceed as follows:

1. Finding the y-intercept:

The y-intercept occurs when $x = 0$.

Substitute $x = 0$ into the equation: $y = \log(3(0) + 4) + 2 = \log(4) + 2$

Now calculate $\log(4)$ using base 10: $\log(4) \approx 0.6021$

So, $y \approx 0.6021 + 2 = 2.6021$

Rounded to the nearest hundredth, the y-intercept is: $y \approx 2.60$

2. Finding the x-intercept:

The x-intercept occurs when $y = 0$.

Set $y = 0$ in the equation: $0 = \log(3x + 4) + 2$

To isolate the logarithmic term, subtract 2 from both sides: $-2 = \log(3x + 4)$

Convert the logarithmic equation to its exponential form: $10^{-2} = 3x + 4$

Now solve for $x$: $0.01 = 3x + 4$ $3x = 0.01 - 4 = -3.99$ $x = \frac{-3.99}{3} \approx -1.33$

Rounded to the nearest hundredth, the x-intercept is: $x \approx -1.33$

Final Answers:

• x-intercept $\approx -1.33$
• y-intercept $\approx 2.60$

Would you like more details on any part of the solution?

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5 Related Questions:

1. How do you find intercepts for exponential functions?
2. What is the general process for converting logarithmic equations to exponential form?
3. How does changing the base of a logarithm affect the intercepts?
4. What happens to the graph if we change the equation to $y = \log(3x + 4) - 2$?
5. Can you explain how to find the intercepts of a function with a natural logarithm?

Tip: When solving for intercepts, always remember that the y-intercept occurs where $x = 0$ and the x-intercept occurs where $y = 0$.