jika 2log 5=a dan 3log 2=b, tentukan 6 log 50?

jika ²log 5 = a ---> [tex] ^{5} log 2 = \frac{1}{5} [/tex]
      ³log 2 = b ---> ²log 3 = 1/b

[tex] ^{6} log 50[/tex]
[tex] ^{6} log 2( 5^{2}) [/tex]
[tex] ^{6} log 2 + ^{6} log 5^{2} [/tex][tex] \frac{1}{ ^{2} log 6 } + 2. ^{6} log 5 [/tex]

[tex] \frac{1}{ ^{2} log (2)(3) } + \frac{2}{ ^{5} log (2)(3) } [/tex]

[tex] \frac{1}{ ^{2} log 2 + ^{2} log 3 } + \frac{2}{ ^{5}log 2 + ^{5} log 3 } [/tex]

[tex] \frac{1}{ ^{2}log 2 + ^{2} log 3 } + \frac{2}{ ^{5} log 2 + ^{5} log 2. ^{2} log 3 } [/tex]

[tex] \frac{1}{1 + \frac{1}{b} } + \frac{2}{ \frac{1}{a} + ( \frac{1}{a})( \frac{1}{b}) } [/tex] samakan penyebutnya

[tex] \frac{1}{ \frac{b + 1}{b} } + \frac{2}{ \frac{b + 1}{ab} } [/tex]

[tex] \frac{b}{b + 1} + \frac{2ab}{b + 1} [/tex]

hasilnya
[tex] \frac{2ab + b}{b + 1} [/tex] sederhanakan menjadi [tex] \frac{b(2a + 1)}{b + 1} [/tex]