Question-9977

Question Number 9977 by konen last updated on 20/Jan/17

Answered by sandy_suhendra last updated on 20/Jan/17

\log_{3^{2}} {(x - 6)}^{2} = \log_{9} (x - 8)

{(x - 6)}^{2} = x - 8

x^{2} - 12 x + 36 = x - 8

x^{2} - 13 x + 44 = 0

x_{1, 2} = \frac{13 \pm \sqrt{169 - 176}}{2} \Rightarrow x = imaginary

Commented by prakash jain last updated on 20/Jan/17

If the question was

\log_{9} (x - 6) = \log_{3} (x - 8)

(x - 6) = {(x - 8)}^{2}

x - 6 = x^{2} - 16 x + 64

x^{2} - 17 x + 70 = 0

(x - 7) (x - 10) = 0

x = 10 [x = 7 does not satisfy]