What is log_{10}(c((a(t)L(t))/(K(t)))^{1-a}) ?

The solution to log_{10}(c((a(t)L(t))/(K(t)))^{1-a}) is log_{10}(c)+log_{10}((Lat)/K)(1-a)

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simplify log_{10}(c((a(t)L(t))/(K(t)))^{1-a})

simplify

lo g_{10} (c (\frac{a ( t ) L ( t )}{K ( t )})^{1 - a})

lo g_{10} (c) + lo g_{10} (\frac{L a t}{K}) (1 - a)

Solution steps

lo g_{10} (c (\frac{a ( t ) L ( t )}{K ( t )})^{1 - a})

Apply log rule:

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

lo g_{10} (c (\frac{a ( t ) L ( t )}{K ( t )})^{1 - a}) = lo g_{10} (c) + lo g_{10} ((\frac{a ( t ) L ( t )}{K ( t )})^{1 - a})

= lo g_{10} (c) + lo g_{10} ((\frac{L a ( t ) ( t )}{K ( t )})^{- a + 1})

Apply log rule:

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

lo g_{10} ((\frac{a ( t ) L ( t )}{K ( t )})^{1 - a}) = (1 - a) lo g_{10} (\frac{a t L t}{K t})

= lo g_{10} (c) + (1 - a) lo g_{10} (\frac{a t L t}{K t})

\frac{a t L t}{K t} = \frac{L a t}{K}

= lo g_{10} (c) + lo g_{10} (\frac{L a t}{K}) (- a + 1)

4 lo g_{2} (x) - 6 lo g_{2} (y)

ln (18 \cdot 1 0^{2})

41.86 \cdot ln (\frac{288}{273})

\frac{1}{3} ln (\frac{10}{3}) \cdot 15 + ln (\frac{1}{10})

ln^{2} (t) + ln (t)

What is log_{10}(c((a(t)L(t))/(K(t)))^{1-a}) ?
The solution to log_{10}(c((a(t)L(t))/(K(t)))^{1-a}) is log_{10}(c)+log_{10}((Lat)/K)(1-a)