Question

At what integer value will the value of y=3(2)^x surpass the value of y=5x^2 +3?

Answer 1

To find the integer value of \( x \) at which the equation \( y = 3(2^x) \) surpasses \( y = 5x^2 + 3 \), we need to solve the inequality:

\[
3(2^x) > 5x^2 + 3
\]

First, we can simplify this to:

\[
3(2^x) - 5x^2 > 3
\]

Next, we can rearrange it as follows:

\[
3(2^x) > 5x^2 + 3
\]

To find a suitable \( x \), we will test integer values of \( x \).

1. **For \( x = 0 \)**:

\[
y = 3(2^0) = 3 \quad \text{and} \quad y = 5(0^2) + 3 = 3
\]

Here, \( 3 \not> 3 \).

2. **For \( x = 1 \)**:

\[
y = 3(2^1) = 6 \quad \text{and} \quad y = 5(1^2) + 3 = 8
\]

Here, \( 6 \not> 8 \).

3. **For \( x = 2 \)**:

\[
y = 3(2^2) = 12 \quad \text{and} \quad y = 5(2^2) + 3 = 23
\]

Here, \( 12 \not> 23 \).

4. **For \( x = 3 \)**:

\[
y = 3(2^3) = 24 \quad \text{and} \quad y = 5(3^2) + 3 = 48
\]

Here, \( 24 \not> 48 \).

5. **For \( x = 4 \)**:

\[
y = 3(2^4) = 48 \quad \text{and} \quad y = 5(4^2) + 3 = 83
\]

Here, \( 48 \not> 83 \).

6. **For \( x = 5 \)**:

\[
y = 3(2^5) = 96 \quad \text{and} \quad y = 5(5^2) + 3 = 128
\]

Here, \( 96 \not> 128 \).

7. **For \( x = 6 \)**:

\[
y = 3(2^6) = 192 \quad \text{and} \quad y = 5(6^2) + 3 = 183
\]

Here, \( 192 > 183 \).

Thus, the integer value of \( x \) at which \( 3(2^x) \) surpasses \( 5x^2 + 3 \) is:

\[
\boxed{6}
\]