Question

What is the value of x in the equation below 1+2e^x+1=9

Answer 1

Answer:

$x=1.2528$

Step-by-step explanation:

Assuming the equation to solve is

$1+2e^x+1=9$

We can first simplify as:

$1+2e^x+1=9\\2+2e^x=9\\2e^x=9-2\\2e^x=7\\e^x=\frac{7}{2}$

to solve an equation with e and x as an exponent, we need to take "natural log (ln)" on both sides and also use the property:

$ln(a^x)=xln(a)$

And also remember that ln e = 1

Now we have:

$e^x=\frac{7}{2}\\ln(e^x)=ln(\frac{7}{2})\\xln(e)=ln(\frac{7}{2})\\x(1)=ln(\frac{7}{2})\\x=ln(\frac{7}{2})\\x=1.2528$

Answer 2

Answer:  $x$≈$1.252$

Step-by-step explanation:

Given the equation $1+2e^x+1=9$, add the like terms:

$2e^x+2=9$

Subtract 2 from both sides:

$2e^x+2-2=9-2$

$2e^x=7$

Divide both sides by 2:

$\frac{2e^x}{2}=\frac{7}{2}\\\\e^x=\frac{7}{2}$

Apply natural logarithm to both sides. Remember that

$ln(e)=1$ and  $ln(m)^n=nln(m)$

Then, you get:

$ln(e)^x=ln(\frac{7}{2})\\\\xln(e)=ln(\frac{7}{2})\\\\x=ln(\frac{7}{2})$

$x$≈$1.252$