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#### DeusAbscondus

##### Active member

- Jun 30, 2012

- 176

(thx kindly: I'm revising stuff I tried to cram last year)

The set question:

Solve for x:

$$y=ln(x)+1$$

Answer given in text:

$$y-1=ln(x)$$

$$\therefore \text {by definition}\ x=e^{y-1}$$

$\text{My attempt, using log laws: }$

$$y=ln(x)+1$$

$$\Rightarrow y=ln(x)+ln(e)$$

$$\Rightarrow y=ln(ex)$$

$$\therefore \text{ by definition} \ e^{y}=ex $$

$$\Rightarrow x=\frac{e^y}{e}$$

$$\therefore x=e^{y-1}$$