How do you solve #e^x + 3 = 6#?

1 Answer
May 31, 2016

#color(green)(x=ln(3) ~~1.099" to 3 decimal places")#

Explanation:

#color(blue)("Introduction of concepts"#

Example of principle: Try this on your calculator

Using log to base 10 enter #log(10)# and you get the answer of 1.

Log to base e is called 'natural' logs and is written as #ln(x)# for any value #x#

#color(brown)("Consequently "ln(e)=1)# Try that on your calculator

[ you may have to enter #ln(e^1)# ]

Another trick is that #log(x^2) -> 2log(x) => ln(x^2)=2ln(x)#

Combining these two ideas:

#ln(e^2)" "=" "2ln(e)" "=" "2xx1=2#

#color(brown)("So "ln(e^x)" "=" "xln(e)" "=" "x xx1" " =" " x)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Solving the question")#

Given:#" "e^x+3=6#

Subtract 3 from both sides

#" "e^x=6-3#

#" "e^x=3#

Take logs of both sides

#" "ln(e^x)=ln(3)#

#" "xln(e)=ln(3)#

But #ln(e)=1# giving

#color(green)(x=ln(3) ~~1.099" to 3 decimal places")#