Broj Klarinih godina jednak je log 5 3 a 2 \log_{\sqrt[3]{5}}a^2 log 3 5 a 2 log 5 5 ( 125 b 9 ) \log_{5\sqrt5}(125b^9) log 5 5 ( 125 b 9 ) 1 log c 2 5 3 \dfrac1{\log_{c^2}\sqrt[3]{5}} log c 2 3 5 1 a b c = 625 abc=625 ab c = 625
Rješenje Zapišimo dane logaritme u jednostavnijem obliku:
log 5 3 a 2 = 2 log 5 1 / 3 a = 6 log 5 a , \log_{\sqrt[3]{5}}a^2=2\log_{5^{1/3}}a=6\log_5a, log 3 5 a 2 = 2 log 5 1/3 a = 6 log 5 a , log 5 5 ( 125 b 9 ) = log 5 3 / 2 125 + log 5 3 / 2 b 9 = 3 ⋅ 2 3 log 5 5 + 9 ⋅ 2 3 log 5 b = 2 + 6 log 5 b , \log_{5\sqrt5}(125b^9)=\log_{5^{3/2}}125+\log_{5^{3/2}}b^9=3\cdot\frac23\log_55+9\cdot\frac23\log_5b=2+6\log_5b, log 5 5 ( 125 b 9 ) = log 5 3/2 125 + log 5 3/2 b 9 = 3 ⋅ 3 2 log 5 5 + 9 ⋅ 3 2 log 5 b = 2 + 6 log 5 b , 1 log c 2 5 3 = 1 1 2 log c 5 1 / 3 = 1 1 6 log c 5 = 6 log 5 c . \frac1{\log_{c^2}\sqrt[3]{5}}=\frac1{\frac12\log_c5^{1/3}}=\frac1{\frac16\log_c5}=6\log_5c. log c 2 3 5 1 = 2 1 log c 5 1/3 1 = 6 1 log c 5 1 = 6 log 5 c . Traženi zbroj godina jednak je:
6 log 5 a + 2 + 6 log 5 b + 6 log 5 c = 2 + 6 log 5 a b c = 2 + 6 log 5 625 = 2 + 6 ⋅ 4 = 26 godina . 6\log_5a+2+6\log_5b+6\log_5c=2+6\log_5abc=2+6\log_5625=2+6\cdot4=26\text{ godina}. 6 log 5 a + 2 + 6 log 5 b + 6 log 5 c = 2 + 6 log 5 ab c = 2 + 6 log 5 625 = 2 + 6 ⋅ 4 = 26 godina . 