Question 239310
Factoring out the minus -1 was a good idea. But, as you found, separating log(-1) was not. With {{{y = log(((-1)(x-8)))}}}, the transformation from {{{y = log((x))}}} is:<ul><li>A translation to the right of 8</li><li>A reflection in the y-axis</li></ul>
Another way could be to factor out -1:
{{{y = log(((-1)(x- 8)))}}}
Factor the -1 into 10*(-1/10):
{{{y = log(((10)(-1/10)(x- 8)))}}}
Separate out the factor of 10:
{{{y = log((10)) + log(((-1/10)(x- 8)))}}}
Since {{{log((10)) = 1}}}:
{{{y = 1 + log(((-1/10)(x- 8)))}}}
The transformations from {{{y = log((x))}}} would be:<ul><li>A translation to the right of 8</li><li>A translation up of 1</li><li>A reflection in the y-axis (because of the "-")</li><li>A horizontal spreading or stretching by a factor of 10 (Stretching, not compression, because of the fraction)</li></ul>
I don't know if this second one is "very different" enough.