- #1
DeusAbscondus
- 176
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Would some kind soul please look over the following and check that use of the log rules, thought roundabout, is nonetheless correct?
(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:
$$y-1=ln(x)$$
$$\therefore \text {by definition}\ x=e^{y-1}$$
$\text{My attempt, using log laws: }$
$$y=ln(x)+1$$
$$\Rightarrow y=ln(x)+ln(e)$$
$$\Rightarrow y=ln(ex)$$
$$\therefore \text{ by definition} \ e^{y}=ex $$
$$\Rightarrow x=\frac{e^y}{e}$$
$$\therefore x=e^{y-1}$$
(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:
$$y-1=ln(x)$$
$$\therefore \text {by definition}\ x=e^{y-1}$$
$\text{My attempt, using log laws: }$
$$y=ln(x)+1$$
$$\Rightarrow y=ln(x)+ln(e)$$
$$\Rightarrow y=ln(ex)$$
$$\therefore \text{ by definition} \ e^{y}=ex $$
$$\Rightarrow x=\frac{e^y}{e}$$
$$\therefore x=e^{y-1}$$
