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  • 18-06-2014
  • Mathematics
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how to solve 2^(log(2)3+ log(2)5

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konrad509
konrad509 konrad509
  • 18-06-2014
[tex]2^{\log_23+ \log_25 }=\\ 2^{\log_215}=\\ 15[/tex]
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Аноним Аноним
  • 20-06-2014
[tex]2^{log_23+log_25}=2^{log_2(3\cdot5)}=2^{log_215}=15\\\\---------------------------\\\\for\ a,b,c\in\mathbb{R^+}\\\\log_ab+log_ac=log_a(b\cdot c)\\\\a^{log_ab}=b[/tex]
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