How do you solve $log (4 p - 2) = log (- 5 p + 5)$ $log (4p-2)=log(-5p+5)$ ?

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Answer 1 · 2018-04-05T14:37:15

$\frac{7}{9}$ $color(blue)(7/9)$

If:

$log (4 p - 2) = log (- 5 p + 5) \Rightarrow 4 p - 2 = - 5 p + 5$ $log(4p-2)=log(-5p+5)=>4p-2=-5p+5$

$4 p - 2 = - 5 p + 5$ $4p-2=-5p+5$

$9 p = 7$ $9p=7$

$p = \frac{7}{9}$ $p=7/9$

Check:

$log (4 (\frac{7}{9}) - 2) = log (- 5 (\frac{7}{9}) + 5)$ $log(4(7/9)-2)=log(-5(7/9)+5)$

$log (\frac{10}{9}) = log (\frac{10}{9})$ $log(10/9)=log(10/9)$

How do you solve log(4p−2)=log(−5p+5)log (4p-2)=log(-5p+5)?