Question

solve. round your answer to the nearest thousandth.\\(2^x + 2 = 3\\)\\(x = \square\\)

Explanation:

Step1: Isolate the exponential term

Subtract 2 from both sides of the equation \(2^x + 2 = 3\) to get \(2^x=3 - 2\), so \(2^x = 1\).

Step2: Take logarithm on both sides

Take the logarithm (we can use natural logarithm or common logarithm, here we use natural logarithm) of both sides: \(\ln(2^x)=\ln(1)\).
Using the logarithm power rule \(\ln(a^b)=b\ln(a)\), we get \(x\ln(2)=\ln(1)\).
Since \(\ln(1) = 0\), the equation becomes \(x\ln(2)=0\). But wait, actually when we had \(2^x = 1\), we know that any non - zero number to the power of 0 is 1, so \(x = 0\). Let's check the process again. Wait, the original equation is \(2^x+2 = 3\), so \(2^x=3 - 2=1\). And we know that \(a^0 = 1\) for \(a>0,a
eq1\), so for \(a = 2\), \(2^0=1\), so \(x = 0\). But if we use the logarithm method:
From \(2^x=1\), taking \(\log_2\) on both sides, \(\log_2(2^x)=\log_2(1)\), then \(x=\log_2(1)=0\) (because \(\log_a(1) = 0\) for \(a>0,a
eq1\)).

Wait, maybe I made a mistake in the initial subtraction. Wait, \(2^x+2 = 3\), subtract 2: \(2^x=3 - 2 = 1\). Then \(x = 0\) because \(2^0=1\). But let's verify: when \(x = 0\), \(2^0+2=1 + 2=3\), which satisfies the original equation.

But if we use the general method for exponential equations:

Starting from \(2^x+2 = 3\)

Step1: Isolate \(2^x\)

\(2^x=3 - 2=1\)

Step2: Solve for \(x\)

We can write \(1\) as \(2^0\), so \(2^x=2^0\). Since the exponential function \(y = 2^x\) is one - to - one (it is strictly increasing), if \(2^x=2^0\), then \(x = 0\).

Answer:

\(x = 0.000\) (rounded to the nearest thousandth)