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USM ITT 2010

Solusi USM IT Telkom 2010 Kode 11 – Nomor 2


Bila\ ^{7}log2=a\ dan\ ^{2}log3=b,\ maka\ ^{6}log98=....
(A)\quad \frac{a}{a+b}
(B)\quad \frac{a+2}{b(a+1)}
(C)\quad \frac{a+2}{b+2}
(D)\quad \frac{a+2}{a(b+1)}
(E)\quad \frac{a+2}{b+1}
Jawaban: D

\begin{array}{rcl} ^{6}log98&=&\frac{^{2}log98}{^{2}log6}\\\\&=&\frac{^{2}log(2\times 49)}{^{2}log(2\times 3)}\\\\&=&\frac{^{2}log2+^{2}log7^{2}}{^{2}log2+^{2}log3}\\\\&=&\frac{1+\big(2\times ^{2}log7\big)}{1+b}\\\\&=&\frac{1+\big(2\times \frac{1}{a}\big)}{b+1}\quad \quad \times \frac{a}{a}\\\\&=&\frac{a+2}{a(b+1)}\end{array}

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