The equation $y = 2^x + 1$ defines a relationship between $x$ and $y$. Solving for $x$ requires isolating it. Here's a step-by-step explanation:

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1. Rewrite the equation:

$y = 2^x + 1$

Subtract 1 from both sides: $y - 1 = 2^x$

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2. Take the logarithm of both sides:

To solve for $x$, use the properties of logarithms. Specifically, take the logarithm (log base 2 is ideal) of both sides: $\log_2(y - 1) = \log_2(2^x)$

Using the logarithmic rule $\log_b(a^c) = c \log_b(a)$, this simplifies to: $\log_2(y - 1) = x$

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3. Final solution:

$x = \log_2(y - 1)$

If you don't have a calculator that computes log base 2, you can use the change of base formula: $\log_2(a) = \frac{\log(a)}{\log(2)}$

So the solution becomes: $x = \frac{\log(y - 1)}{\log(2)}$

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Conditions:

• $y > 1$: This is necessary because $y - 1$ must be positive for the logarithm to exist.

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Let me know if you'd like a detailed numerical example or further explanation! 😊

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Related Questions:

1. How does the graph of $y = 2^x + 1$ look compared to $y = 2^x$?
2. What is the domain and range of $y = 2^x + 1$?
3. How do you solve $y = 3^x + 2$ instead?
4. What is the derivative of $y = 2^x + 1$?
5. How can logarithms simplify solving exponential equations in real-life problems?

Tip: When solving exponential equations, remember to check for restrictions in the domain due to the logarithmic function.