To find the inverse of the given exponential function:
[tex]\[ y = \frac{5^{2x+1}}{2} \][/tex]
we will follow these steps:
1. Interchange the roles of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ x = \frac{5^{2y+1}}{2} \][/tex]
2. Solve for [tex]\(y\)[/tex] from the above equation.
Let's solve this step-by-step:
### Step 1: Interchange [tex]\(x\)[/tex] and [tex]\(y\)[/tex]
[tex]\[ x = \frac{5^{2y+1}}{2} \][/tex]
### Step 2: Isolate the exponential term
Multiply both sides of the equation by 2 to get rid of the fraction:
[tex]\[ 2x = 5^{2y+1} \][/tex]
### Step 3: Apply logarithms to both sides
Since the base of the exponent is 5, use the logarithm base 5:
[tex]\[ \log_5(2x) = \log_5(5^{2y+1}) \][/tex]
### Step 4: Use the properties of logarithms
The logarithmic property [tex]\( \log_b(a^c) = c \cdot \log_b(a) \)[/tex] tells us:
[tex]\[ \log_5(2x) = (2y+1) \cdot \log_5(5) \][/tex]
Since [tex]\( \log_5(5) = 1 \)[/tex], we have:
[tex]\[ \log_5(2x) = 2y + 1 \][/tex]
### Step 5: Solve for [tex]\(y\)[/tex]
Isolate [tex]\(y\)[/tex] on one side of the equation:
[tex]\[ \log_5(2x) - 1 = 2y \][/tex]
Divide both sides by 2:
[tex]\[ y = \frac{\log_5(2x) - 1}{2} \][/tex]
Therefore, the inverse of the function [tex]\( y = \frac{5^{2x+1}}{2} \)[/tex] is:
[tex]\[ y^{-1}(x) = \frac{\log_5(2x) - 1}{2} \][/tex]
However, consider that logarithms can have multiple solutions depending on various branches stemming from the behavior of the function values. The complete solution involves complex logarithms which results in:
[tex]\[ [\log(2x/5)/(2\log(5)), \log(-\sqrt{10}\sqrt(x)/5)/\log(5)] \][/tex]
For practical purposes, the principal branch of the solution is often used:
[tex]\[ y^{-1}(x) = \frac{\log_5(2x) - 1}{2} \][/tex]
