What is e^{(y-1)}-ln(y-1)-ln(x) ?

The solution to e^{(y-1)}-ln(y-1)-ln(x) is e^{y-1}-ln((y-1)x)

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simplify e^{(y-1)}-ln(y-1)-ln(x)

simplify

e^{(y - 1)} - ln (y - 1) - ln (x)

e^{y - 1} - ln ((y - 1) x)

Solution steps

e^{(y - 1)} - ln (y - 1) - ln (x)

Rewrite

- ln (y - 1) - ln (x)

- (ln (y - 1) + ln (x))

= e^{(y - 1)} - (ln (y - 1) + ln (x))

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

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

= e^{y - 1} - ln ((y - 1) x)

ln (\frac{x}{x + 2})

ln (\frac{1}{1999}) \cdot 0.499

lo g_{10} (z) + 3 lo g_{10} (y)

- lo g_{10} (1 x 1 0^{- 8})

c^{x} - ln (c^{x}) + ln (x)

What is e^{(y-1)}-ln(y-1)-ln(x) ?
The solution to e^{(y-1)}-ln(y-1)-ln(x) is e^{y-1}-ln((y-1)x)