Jak to rozpisać?1. 2. 3. 4. 5. Nie czaję działań na logarytmach.... Question from @pleso93

September 2018 1 41 Report

Jak to rozpisać?
1. $2^{2\sqrt{3} + log_2 7}$
2. $4^{\sqrt{3} + 1}$
3. $log2*log50+log^{2}5$
4. $(log_{3}36)^{2}-log_{3}16*log_{3}18$
5. $log_{4}\sqrt{5}*log_{25}8$

Nie czaję działań na logarytmach...

Roma

$2^{2\sqrt{3} + log_2 7} = 2^{2\sqrt{3}} \cdot 2^{log_2 7}} = (2^2)^{\sqrt{3}} \cdot 7 = 7 \cdot 4^{\sqrt{3}}$

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$a^{m + n} = a^m \cdot a^n$

$a^{m \cdot n} = (a^m)^n$

$a^{log_a x} = x$

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$4^{\sqrt{3} + 1} = 4^{\sqrt{3}} \cdot 4^1 = 4 \cdot 4^{\sqrt{3}}$

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$a^{m + n} = a^m \cdot a^n$

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$log 2 \cdot log 50+log^2 5 = log 2 \cdot log (5 \cdot 10) + log^2 5 =$

$= log 2 \cdot (log 5 + log 10) + log 5 \cdot log 5 = log 2 \cdot (log 5 + 1) + log 5 \cdot log 5 =$

$= log 2 \cdot log 5 + log 2 + log 5 \cdot log 5 = log 2 \cdot log 5 + log 5 \cdot log 5 + log 2 =$

$= log 5 \cdot (log 2 + log 5) + log 2 = log 5 \cdot log (2 \cdot 5) + log 2 =$

$= log 5 \cdot 1 + log 2 = log 5 + log 2 = log (5 \cdot 2) = log 10 = 1$

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$log_a (b \cdot c) = log_a b + log_a c$

$log 10 = x$

$10^x = 10$

$10^x = 10^1$

$x = 1$

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$(log_3 36)^{2}-log_3 16 \cdot log_318 = log_3 36 \cdot log_3 36-log_3 2^4 \cdot log_3(2 \cdot 9) =$

$= log_3 (4 \cdot 9) \cdot log_3 (4 \cdot 9) - 4log_3 2 \cdot (log_32 + log_3 9) =$

$= (log_3 4+ log_3 9) \cdot (log_3 4+ log_3 9) - 4log_3 2 \cdot (log_32 + 2) =$

$= (log_3 2^2+ 2) \cdot (log_3 2^2+ 2) - (4log_3 2 \cdot log_32 + 8log_3 2) =$

$= (2log_3 2+ 2) \cdot (2log_3 2+ 2) - 4log_3 2 \cdot log_32 - 8log_3 2 =$

$= 4log_3 2 \cdot log_3 2 + 4log_3 2 + 4log_3 2 + 4 - 4log_3 2 \cdot log_32 - 8log_3 2 =$

$= 4$

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$log_a (b \cdot c) = log_a b + log_a c$

$log_a b^r = r \cdot log_a b$

$log_3 9 = x$

$3^x = 9$

$3^x = 3^2$

$x = 2$

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$log_4 \sqrt{5} \cdot log_{25} 8 = \frac{log_5 \sqrt{5}}{log_5 4} \cdot \frac{log_5 8}{log_5 25} = \frac{log_5 5^{\frac{1}{2}}}{log_5 4} \cdot \frac{log_5 (2 \cdot 4)}{log_5 5^2} =$

$= \frac{\frac{1}{2}log_5 5}{log_5 4} \cdot \frac{log_5 2 + log_5 4}{2log_5 5} = \frac{\frac{1}{2}}{log_5 4} \cdot \frac{log_5 2 + log_5 4}{2} = \frac{\frac{1}{2}(log_5 2 + log_5 4)}{2log_5 4} =$

$= \frac{1}{4} \cdot \frac{log_5 2 + log_5 4}{log_5 4} = \frac{1}{4} \cdot (\frac{log_5 2}{log_5 4} + \frac{log_5 4}{log_5 4}) = \frac{1}{4} \cdot (\frac{log_5 2}{log_5 2^2} + 1) =$

$= \frac{1}{4} \cdot (\frac{log_5 2}{2log_5 2} + 1) = \frac{1}{4} \cdot (\frac{1}{2} + 1) = \frac{1}{8} + \frac{1}{4} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$