How do you solve $e^{x} + 3 = 6$ $e^x + 3 = 6$ ?

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Answer 1 · 2016-05-31T18:13:37

$x = ln (3) \approx 1.099 to 3 decimal places$ $color(green)(x=ln(3) ~~1.099" to 3 decimal places")$

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

Example of principle: Try this on your calculator

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

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

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

[ you may have to enter $ln (e^{1})$ $ln(e^1)$ ]

Another trick is that $log (x^{2}) \to 2 log (x) \Rightarrow ln (x^{2}) = 2 ln (x)$ $log(x^2) -> 2log(x) => ln(x^2)=2ln(x)$

Combining these two ideas:

$ln (e^{2}) = 2 ln (e) = 2 \times 1 = 2$ $ln(e^2)" "=" "2ln(e)" "=" "2xx1=2$

$So ln (e^{x}) = x ln (e) = x \times 1 = x$ $color(brown)("So "ln(e^x)" "=" "xln(e)" "=" "x xx1" " =" " x)$

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$Solving the question$ $color(blue)("Solving the question")$

Given: $e^{x} + 3 = 6$ $" "e^x+3=6$

Subtract 3 from both sides

$e^{x} = 6 - 3$ $" "e^x=6-3$

$e^{x} = 3$ $" "e^x=3$

Take logs of both sides

$ln (e^{x}) = ln (3)$ $" "ln(e^x)=ln(3)$

$x ln (e) = ln (3)$ $" "xln(e)=ln(3)$

But $ln (e) = 1$ $ln(e)=1$ giving

$x = ln (3) \approx 1.099 to 3 decimal places$ $color(green)(x=ln(3) ~~1.099" to 3 decimal places")$

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