Question

solve for x to the nearest 10th 0=4^x+1 -2

Answer 1

To solve the equation \( 0 = 4^{x+1} - 2 \), we first isolate the exponential term:

\[
4^{x+1} = 2
\]

Next, we can express \( 4 \) as \( 2^2 \):

\[
(2^2)^{x+1} = 2
\]

This simplifies to:

\[
2^{2(x+1)} = 2
\]

Since \( 2 \) can be written as \( 2^1 \), we have:

\[
2^{2(x+1)} = 2^1
\]

Since the bases are the same, we can equate the exponents:

\[
2(x+1) = 1
\]

Now, we solve for \( x \):

\[
2x + 2 = 1
\]

Subtract \( 2 \) from both sides:

\[
2x = -1
\]

Divide by \( 2 \):

\[
x = -\frac{1}{2}
\]

To the nearest tenth, \( x = -0.5 \).

Thus, the solution is:

\[
\boxed{-0.5}
\]