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Solve [tex]$e^{2x+1} = 33$[/tex] for [tex]$x$[/tex].

[tex]x =[/tex]


Sagot :

To solve the equation [tex]\( e^{2x + 1} = 33 \)[/tex] for [tex]\( x \)[/tex], we can follow these steps:

1. Isolate the exponential term:

Given the equation [tex]\( e^{2x + 1} = 33 \)[/tex], we want to isolate the exponential expression. We'll begin by taking the natural logarithm on both sides of the equation to eliminate the exponential function. The natural logarithm ([tex]\( \ln \)[/tex]) and the exponential function ([tex]\( e \)[/tex]) are inverse functions.

2. Apply the natural logarithm:

[tex]\[ \ln(e^{2x + 1}) = \ln(33) \][/tex]

3. Simplify the left side:

Using the property of logarithms that [tex]\( \ln(e^y) = y \)[/tex], we simplify the left-hand side:

[tex]\[ 2x + 1 = \ln(33) \][/tex]

4. Isolate [tex]\( x \)[/tex]:

Rearrange the equation to solve for [tex]\( x \)[/tex]. First, subtract 1 from both sides:

[tex]\[ 2x = \ln(33) - 1 \][/tex]

Then, divide both sides by 2:

[tex]\[ x = \frac{\ln(33) - 1}{2} \][/tex]

Simplifying this expression:

[tex]\[ x = \frac{\ln(33) - 1}{2} \][/tex]

So, the solution to the equation [tex]\( e^{2x + 1} = 33 \)[/tex] for [tex]\( x \)[/tex] is:

[tex]\[ x = -\frac{1}{2} + \frac{\ln(33)}{2} \][/tex]