What is expand-1(-log_{4}(x-3)=-2) ?

The solution to expand-1(-log_{4}(x-3)=-2) is log_{4}(x-3)=2

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expand-1(-log_{4}(x-3)=-2)

expand

- 1 (- lo g_{4} (x - 3) = - 2)

lo g_{4} (x - 3) = 2

Solution steps

- 1 \cdot (- lo g_{4} (x - 3) = - 2)

= - 1 \cdot (- = lo g_{4} (x - 3) - 2)

Apply the distributive law:

a (b - c) = ab - a c

a = - 1, b = - lo g_{4} (x - 3) =, c = 2

= - 1 \cdot (- lo g_{4} (x - 3) =) - (- 1) \cdot 2

Apply minus-plus rules

- (- a) = a

1 \cdot lo g_{4} (x - 3) = 1 \cdot 2

Simplify

1 \cdot lo g_{4} (x - 3) = 1 \cdot 2 : lo g_{4} (x - 3) = 2

lo g_{4} (x - 3) = 2

\frac{2 y ^{3} - 21 y ^{2} + 102 y + 75}{2 ( y + 5 ) ( y - 6 )}

- \frac{2}{( x - 2 ) ^{2} ( x + 2 )}

(b + f) (e j c + e (a f + a + f + 1 + e - e b))

\frac{3 h}{( a + 1 ) ( h + a + 1 )}

(3 x + 1 0^{\circ}) (2 x - 5^{\circ}) (x - 5^{\circ})

What is expand-1(-log_{4}(x-3)=-2) ?
The solution to expand-1(-log_{4}(x-3)=-2) is log_{4}(x-3)=2