Log2(3)=a log2(5)=blog4(135)=?​

$\log_{2}3 = a, \ \log_{2}5 = b$

$\log_{4}135 = \log_{2^{2}}(27 \cdot 5) = \dfrac{1}{2} \log_{2}3^{3} + \dfrac{1}{2}\log_{2}5 =\dfrac{3}{2} \log_{2}3 + \dfrac{1}{2}\log_{2}5 = \dfrac{3}{2} a + \dfrac{1}{2} b$

Свойства логарифмов:

0, \ a \neq 1, \ b > 0)" alt="\log_{a^{p}}b = \dfrac{1}{p}\log_{a}b \ \ \ (a > 0, \ a \neq 1, \ b > 0)" align="absmiddle" class="latex-formula">

0, \ a \neq 1, \ b > 0)" alt="\log_{a}b^{p} = p\log_{a}b \ \ \ (a > 0, \ a \neq 1, \ b > 0)" align="absmiddle" class="latex-formula">

0, \ a\neq 1, \ x > 0, \ y > 0)" alt="\log_{a}(xy) = \log_{a}x + \log_{a}y \ (a>0, \ a\neq 1, \ x > 0, \ y > 0)" align="absmiddle" class="latex-formula">

Log2(3)=a log2(5)=blog4(135)=?​

Log2(3)=a log2(5)=blog4(135)=?