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yulamanovaguldar
@yulamanovaguldar
July 2022
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Помогите пожалуйста - Прологарифмируйте данное выражение по основанию а:
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sangers1959
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1) logₐ A=logₐx⁴.
2) logₐA=logₐ(a²*2x³y⁴)=logₐa²+log(2x³y⁴)=2+logₐ(2x³y⁴).
3) logₐA=logₐ(∛a²*b*√c/(3*√3*d))=logₐa²/³*(b*√c/(3*√3*d)=
=2/3+logₐ(b*√c/(3*√3*d)=2/3+logₐ(b*√c)-logₐ(3*√3*d).
4) logₐA=logₐ(3*√a/(a*b))=logₐ(a⁻¹/²*(3/b))=-1/2-logₐ(3/b).
5) logₐA=logₐ(a²/(a²/³*b¹/⁴*c¹/²))=logₐ(a⁴/³)/(b¹/⁴*c¹/²)=4/3-logₐ(b¹/⁴*c¹/²).
6) logₐA=logₐ(a³*(a-2)*(a-5))=logₐa³+logₐ(a-2)+logₐ(a-5)=3+logₐ(a-2)+logₐ(a-5).
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yulamanovaguldar
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sangers1959
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Answers & Comments
Verified answer
1) logₐ A=logₐx⁴.2) logₐA=logₐ(a²*2x³y⁴)=logₐa²+log(2x³y⁴)=2+logₐ(2x³y⁴).
3) logₐA=logₐ(∛a²*b*√c/(3*√3*d))=logₐa²/³*(b*√c/(3*√3*d)=
=2/3+logₐ(b*√c/(3*√3*d)=2/3+logₐ(b*√c)-logₐ(3*√3*d).
4) logₐA=logₐ(3*√a/(a*b))=logₐ(a⁻¹/²*(3/b))=-1/2-logₐ(3/b).
5) logₐA=logₐ(a²/(a²/³*b¹/⁴*c¹/²))=logₐ(a⁴/³)/(b¹/⁴*c¹/²)=4/3-logₐ(b¹/⁴*c¹/²).
6) logₐA=logₐ(a³*(a-2)*(a-5))=logₐa³+logₐ(a-2)+logₐ(a-5)=3+logₐ(a-2)+logₐ(a-5).