Solve the equation 2^x = 32.

The solution to the equation 2^x = 32 is x = 5.

To solve this equation, we need to use logarithms. Specifically, we need to take the logarithm of both sides of the equation with base 2, since we are trying to solve for x in terms of 2. This gives us:

log2(2^x) = log2(32)

Using the property of logarithms that states loga(b^c) = c*loga(b), we can simplify the left-hand side of the equation:

x*log2(2) = log2(32)

Since log2(2) = 1, we can simplify further:

x = log2(32)

Now we just need to evaluate log2(32). We can do this by using the change of base formula, which states that loga(b) = logc(b)/logc(a). In this case, we can use either natural logarithms (ln) or common logarithms (log10) as our new base. Let's use ln:

x = ln(32)/ln(2)

Using a calculator, we find that ln(32) ≈ 3.4657 and ln(2) ≈ 0.6931, so:

x ≈ 3.4657/0.6931 ≈ 4.9999

Since x must be an integer (since we are dealing with exponents), we round up to the nearest integer to get x = 5. Therefore, the solution to the equation 2^x = 32 is x = 5.

To deepen your understanding of how logarithms are used to solve equations, you may explore further details on logarithmic functions. Additionally, for a focused study on solving logarithmic equations, consider reviewing the concepts on solving logarithmic equations.

A-Level Maths Tutor Summary: To solve 2^x = 32, we use logarithms to change the equation to x = log2(32). By applying the change of base formula with natural logarithms, we find x ≈ 4.9999. Since x represents an exponent and must be an integer, we round it to the nearest whole number, giving us x = 5. This method demonstrates how logarithms can solve exponent equations. For more on the manipulation of logarithmic and exponential functions, refer to the page on differentiation of exponential and logarithmic functions.