What is the answer to 7=alog_{1/3}(4x-8)-3 ?

The answer to 7=alog_{1/3}(4x-8)-3 is x=((1/3)^{(10)/a}+8)/(4)

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7=alog_{1/3}(4x-8)-3

7 = a lo g_{\frac{1}{3}} (4 x - 8) - 3

x = \frac{( \frac{1}{3} ) ^{\frac{10}{a}} + 8}{4}

Solution steps

7 = a lo g_{\frac{1}{3}} (4 x - 8) - 3

Switch sides

a lo g_{\frac{1}{3}} (4 x - 8) - 3 = 7

Move

3

to the right side

a lo g_{\frac{1}{3}} (4 x - 8) = 10

Divide both sides by

a

lo g_{\frac{1}{3}} (4 x - 8) = \frac{10}{a}

Apply log rules

4 x - 8 = (\frac{1}{3})^{\frac{10}{a}}

Solve

4 x - 8 = (\frac{1}{3})^{\frac{10}{a}} : x = \frac{( \frac{1}{3} ) ^{\frac{10}{a}} + 8}{4}

x = \frac{( \frac{1}{3} ) ^{\frac{10}{a}} + 8}{4}

Verify Solutions:

x = \frac{( \frac{1}{3} ) ^{\frac{10}{a}} + 8}{4}

True

The solution is

x = \frac{( \frac{1}{3} ) ^{\frac{10}{a}} + 8}{4}

x ln (x) + x = 0

lo g_{8} (25) = 2 lo g_{8} (x)

ln (x^{6}) - ln (x^{3}) = 3

2 ln (x) + 2 = 0

x, ln (y - x^{2}) = 0

What is the answer to 7=alog_{1/3}(4x-8)-3 ?
The answer to 7=alog_{1/3}(4x-8)-3 is x=((1/3)^{(10)/a}+8)/(4)