How do you solve $log (4 k - 5) = log (2 k - 1)$ $log (4k-5)=log(2k-1)$ ?

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How do you solve $log (4 k - 5) = log (2 k - 1)$ $log (4k-5)=log(2k-1)$ ?

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Answer 1 · 2016-11-18T18:58:58

$k = 2$ $k=2$

Explanation:

Using the $laws of logarithms$ $color(blue)"laws of logarithms"$

$color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(logx=logyrArrx=y)color(white)(2/2)|)))$

$rArr4k-5=2k-1$

$2k=4rArrk=2$

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How do you solve $log (4 k - 5) = log (2 k - 1)$ $log (4k-5)=log(2k-1)$ ?

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How do you solve log (4k-5)=log(2k-1)log(4k−5)=log(2k−1)log (4k-5)=log(2k-1)?

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How do you solve $log (4 k - 5) = log (2 k - 1)$ $log (4k-5)=log(2k-1)$ ?