What is 5log_{5}(5z+1)+1/2 log_{5}(z+8) ?

The solution to 5log_{5}(5z+1)+1/2 log_{5}(z+8) is log_{5}((5z+1)^5(z+8)^{1/2})

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simplify 5log_{5}(5z+1)+1/2 log_{5}(z+8)

simplify

5 lo g_{5} (5 z + 1) + \frac{1}{2} lo g_{5} (z + 8)

lo g_{5} ((5 z + 1)^{5} (z + 8)^{\frac{1}{2}})

Solution steps

5 lo g_{5} (5 z + 1) + \frac{1}{2} lo g_{5} (z + 8)

Apply log rule:

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

5 lo g_{5} (5 z + 1) = lo g_{5} ((5 z + 1)^{5})

= lo g_{5} ((5 z + 1)^{5}) + \frac{1}{2} lo g_{5} (z + 8)

Apply log rule:

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

\frac{1}{2} lo g_{5} (z + 8) = lo g_{5} ((z + 8)^{\frac{1}{2}})

= lo g_{5} ((5 z + 1)^{5}) + lo g_{5} ((z + 8)^{\frac{1}{2}})

Apply log rule:

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

lo g_{5} ((5 z + 1)^{5}) + lo g_{5} ((z + 8)^{\frac{1}{2}}) = lo g_{5} ((5 z + 1)^{5} (z + 8)^{\frac{1}{2}})

= lo g_{5} ((5 z + 1)^{5} (z + 8)^{\frac{1}{2}})

ln (2 e^{- x^{2}} x)

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\frac{2 ln ( 2 )}{2 ln ( 2 ) - ln ( 3 )}

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What is 5log_{5}(5z+1)+1/2 log_{5}(z+8) ?
The solution to 5log_{5}(5z+1)+1/2 log_{5}(z+8) is log_{5}((5z+1)^5(z+8)^{1/2})