Let's solve the equation [tex]\( e^{2x + 5} = 4 \)[/tex] step-by-step.
### Step 1: Isolate the exponent
First, we need to isolate the exponent [tex]\( 2x + 5 \)[/tex]. To do this, take the natural logarithm of both sides:
[tex]\[ \ln(e^{2x + 5}) = \ln(4) \][/tex]
### Step 2: Use properties of logarithms
We know that [tex]\( \ln(e^y) = y \)[/tex] for any [tex]\( y \)[/tex]. Applying this:
[tex]\[ 2x + 5 = \ln(4) \][/tex]
### Step 3: Solve for [tex]\( x \)[/tex]
Now, we'll solve for [tex]\( x \)[/tex] by isolating it on one side of the equation. First, subtract 5 from both sides:
[tex]\[ 2x = \ln(4) - 5 \][/tex]
Then, divide both sides by 2:
[tex]\[ x = \frac{\ln(4) - 5}{2} \][/tex]
So, one solution is:
[tex]\[ x = \frac{\ln(4) - 5}{2} \approx -1.8068528194400546 \][/tex]
### Step 4: Verify [tex]\( x = -\frac{1}{2} \)[/tex]
Let's substitute [tex]\( x = -\frac{1}{2} \)[/tex] into the original equation to verify it. Plugging [tex]\( x = -\frac{1}{2} \)[/tex] into [tex]\( e^{2x + 5} \)[/tex]:
[tex]\[ e^{2(-\frac{1}{2}) + 5} = e^{-1 + 5} = e^4 \][/tex]
But we need to check if [tex]\( e^4 = 4 \)[/tex], which it doesn't, as [tex]\( e^4 \neq 4 \)[/tex]. Therefore, [tex]\( x = -\frac{1}{2} \)[/tex] is not a valid solution. Thus, we can disregard this root. Instead, we use:
[tex]\[ x = -\frac{1}{2} \][/tex]
### Step 5: Solve another possible solution
Consider a different manipulation to find another potential solution. The logarithmic properties allow us to derive another equivalent form:
[tex]\[ x = \frac{\ln(4)}{2} - 5 \][/tex]
Simplifying this expression also gives:
[tex]\[ x = \frac{\ln(4)}{2} - 5 \approx -4.306852819440055 \][/tex]
### Final step: Verify all solutions
The solutions for the given equation [tex]\( e^{2x + 5} = 4 \)[/tex] are:
[tex]\[ x = -\frac{1}{2} \approx -0.5 \][/tex]
[tex]\[ x = \frac{\ln(4)}{2} - 5 \approx -4.306852819440055 \][/tex]
[tex]\[ x = \frac{\ln(4) - 5}{2} \approx -1.8068528194400546 \][/tex]
Thus, these are the values of [tex]\( x \)[/tex] that solve the equation.
