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

May 31, 2016

$\textcolor{g r e e n}{x = \ln \left(3\right) \approx 1.099 \text{ to 3 decimal places}}$

#### Explanation:

color(blue)("Introduction of concepts"

Example of principle: Try this on your calculator

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

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

$\textcolor{b r o w n}{\text{Consequently } \ln \left(e\right) = 1}$ Try that on your calculator

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

Another trick is that $\log \left({x}^{2}\right) \to 2 \log \left(x\right) \implies \ln \left({x}^{2}\right) = 2 \ln \left(x\right)$

Combining these two ideas:

$\ln \left({e}^{2}\right) \text{ "=" "2ln(e)" "=" } 2 \times 1 = 2$

$\textcolor{b r o w n}{\text{So "ln(e^x)" "=" "xln(e)" "=" "x xx1" " =" } x}$

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$\textcolor{b l u e}{\text{Solving the question}}$

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

Subtract 3 from both sides

$\text{ } {e}^{x} = 6 - 3$

$\text{ } {e}^{x} = 3$

Take logs of both sides

$\text{ } \ln \left({e}^{x}\right) = \ln \left(3\right)$

$\text{ } x \ln \left(e\right) = \ln \left(3\right)$

But $\ln \left(e\right) = 1$ giving

$\textcolor{g r e e n}{x = \ln \left(3\right) \approx 1.099 \text{ to 3 decimal places}}$