What is 3log_{b}(y)+7log_{b}(z) ?

The solution to 3log_{b}(y)+7log_{b}(z) is log_{b}(y^3z^7)

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simplify 3log_{b}(y)+7log_{b}(z)

simplify

3 lo g_{b} (y) + 7 lo g_{b} (z)

lo g_{b} (y^{3} z^{7})

Solution steps

3 lo g_{b} (y) + 7 lo g_{b} (z)

Apply log rule:

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

3 lo g_{b} (y) = lo g_{b} (y^{3})

= lo g_{b} (y^{3}) + 7 lo g_{b} (z)

Apply log rule:

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

7 lo g_{b} (z) = lo g_{b} (z^{7})

= lo g_{b} (y^{3}) + lo g_{b} (z^{7})

Apply log rule:

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

lo g_{b} (y^{3}) + lo g_{b} (z^{7}) = lo g_{b} (y^{3} z^{7})

= lo g_{b} (y^{3} z^{7})

ln (\frac{1}{23})

9 \cdot ln (8) - ln (2!) + 7 \cdot ln (2) - ln (7!) - 3 \cdot 8

3 lo g_{b} (z) + 2 (lo g_{b} (y) - 5 lo g_{b} (x))

- \frac{( 8.314 \cdot 298 )}{96485} ln (\frac{0.002}{4})

lo g_{\frac{1}{2}} (\frac{1}{80})

What is 3log_{b}(y)+7log_{b}(z) ?
The solution to 3log_{b}(y)+7log_{b}(z) is log_{b}(y^3z^7)