Solve for x log 4 ⁡ ( x − 3 ) = 1 − log 4 ⁡ ( x − 6 )

Question

Solve for x      log    4    ⁡  (  x  −  3  )  =  1  −      log    4    ⁡  (  x  −  6  )

lavarcar2d2 · Accepted Answer

Using the properties of logarithmic functions, it can be written that

\log_{a} b + \log_{a} c = \log_{a} (b \times c)

If

\log_{a} b = c

, then

b = a^{c}

Simplify the equation,

\log_{4} (x - 3) = 1 - l o g_{4} (x - 6)

using the properties of the logarithmic functions. Solve the resulting quadratic equation by factoring.

\log_{4} (x - 3) = 1 - \log_{4} (x - 6) \log_{4} (x - 3) + \log_{4} (x - 6) = 1 \log_{4} [(x - 3) (x - 6)] = 1 (x - 3) (x - 6) = 4^{1} x^{2} - 9 x + 18 = 4 x^{2} - 9 x + 14 = 0 x^{2} - 7 x - 2 x + 14 = 0 x (x - 7) - 2 (x - 7) = 0 (x - 2) (x - 7) = 0 x = 2, 7

The solution, x=2 is discarded as x=2 makes the equation,

\log_{4} (x - 3) = 1 - \log_{4} (x - 6)

invalid. Hence, the solution to the given equation is x=7.

Solve for x log_4 (x-3)=1-log_4 (x-6)