What is expand log_{10}(sqrt(a^6b)) ?

The solution to expand log_{10}(sqrt(a^6b)) is 3log_{10}(a)+1/2 log_{10}(b)

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expand log_{10}(sqrt(a^6b))

expand

lo g_{10} (a^{6} b)

3 lo g_{10} (a) + \frac{1}{2} lo g_{10} (b)

Solution steps

lo g_{10} (a^{6} b)

a^{6} b = a^{3} b

= lo g_{10} (a^{3} b)

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

lo g_{10} (a^{3} b) = lo g_{10} (b) + lo g_{10} (a^{3})

= lo g_{10} (b) + lo g_{10} (a^{3})

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{10} (b) + lo g_{10} (a^{3}) = lo g_{10} (a^{3} b)

= lo g_{10} (b a^{3})

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

lo g_{10} (b a^{3}) = lo g_{10} (a^{3}) + lo g_{10} (b)

= lo g_{10} (a^{3}) + lo g_{10} (b)

Simplify

lo g_{10} (b) : \frac{1}{2} lo g_{10} (b)

= lo g_{10} (a^{3}) + \frac{1}{2} lo g_{10} (b)

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

= 3 lo g_{10} (a) + \frac{1}{2} lo g_{10} (b)

d x (e^{s i n (x)} + cos (x))

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

(3 x + 2 y = 1, 6) (y = - 7 x + 19)

(e^{x} cos (x) - sin (x) e^{x}) (- 2 e^{x} sin (x))

\frac{( x + 1 )}{2 x ( 2 x + 1 )}

What is expand log_{10}(sqrt(a^6b)) ?
The solution to expand log_{10}(sqrt(a^6b)) is 3log_{10}(a)+1/2 log_{10}(b)