What is 4log_{c}(2z+1)+1/5 log_{c}(z+2) ?

The solution to 4log_{c}(2z+1)+1/5 log_{c}(z+2) is log_{c}((2z+1)^4(z+2)^{1/5})

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simplify 4log_{c}(2z+1)+1/5 log_{c}(z+2)

simplify

4 lo g_{c} (2 z + 1) + \frac{1}{5} lo g_{c} (z + 2)

lo g_{c} ((2 z + 1)^{4} (z + 2)^{\frac{1}{5}})

Solution steps

4 lo g_{c} (2 z + 1) + \frac{1}{5} lo g_{c} (z + 2)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

4 lo g_{c} (2 z + 1) = lo g_{c} ((2 z + 1)^{4})

= lo g_{c} ((2 z + 1)^{4}) + \frac{1}{5} lo g_{c} (z + 2)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{1}{5} lo g_{c} (z + 2) = lo g_{c} ((z + 2)^{\frac{1}{5}})

= lo g_{c} ((2 z + 1)^{4}) + lo g_{c} ((z + 2)^{\frac{1}{5}})

Apply log rule:

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

lo g_{c} ((2 z + 1)^{4}) + lo g_{c} ((z + 2)^{\frac{1}{5}}) = lo g_{c} ((2 z + 1)^{4} (z + 2)^{\frac{1}{5}})

= lo g_{c} ((2 z + 1)^{4} (z + 2)^{\frac{1}{5}})

lo g_{1.12} (\frac{125}{12}) + 1

0.3 \cdot ln (3400) + 0.7 \cdot ln (5800)

40 e^{\frac{l n ( 53 ) 31}{21}}

ln (\frac{1}{x ^{2} + 9})

(lo g_{30} (2) + lo g_{30} (3) + lo g_{2} (12))^{2}

What is 4log_{c}(2z+1)+1/5 log_{c}(z+2) ?
The solution to 4log_{c}(2z+1)+1/5 log_{c}(z+2) is log_{c}((2z+1)^4(z+2)^{1/5})