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

Answer 1

Benjamin Achtenberg

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

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

Example of principle: Use your calculator to do this.

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)#

Putting these two concepts together:

#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#

Deduct three from each side.

#" "e^x=6-3#

#" "e^x=3#

Take both sides' logs.

#" "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")#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.