Question Number 221135 by MathematicalUser2357 last updated on 25/May/25
$$\mathrm{South}\:\mathrm{Korean}\:\mathrm{Grade}\:\mathrm{12}\:\mathrm{math} \\ $$$$\mathrm{Prove}\:\mathrm{log}_{{a}} {M}^{{n}} ={n}\mathrm{log}_{{a}} {M} \\ $$$$\mathrm{Using}\:\mathrm{below}: \\ $$$$\mathrm{When}\:{M}={a}^{{x}} ,\:\mathrm{log}_{{a}} {M}={x} \\ $$$$\mathrm{When}\:{N}={a}^{{y}} ,\:\mathrm{log}_{{a}} {N}={y} \\ $$$${MN}={a}^{{x}} ×{a}^{{y}} ={a}^{{x}+{y}} \\ $$$$\mathrm{So},\:\mathrm{log}_{{a}} \left({MN}\right)=\mathrm{log}_{{a}} \left({a}^{{x}+{y}} \right)={x}+{y}=\mathrm{log}_{{a}} {M}+\mathrm{log}_{{a}} {N} \\ $$
Commented by SdC355 last updated on 25/May/25
이지하군요ㅋㅋㅋㅋ
너무 쉬움
Commented by fantastic last updated on 25/May/25
$${I}\:{don}'{t}\:{believe}\:{that}\:{it}\:{is}\:{a}\:{grade}\:\mathrm{12}\:{question} \\ $$
Answered by fantastic last updated on 25/May/25
![Let M,a,n be real numbers and M>0 ,a>0,a≠1 Let log _a M=x So a^x =M. ∴M^n =(a^x )^n =a^(xn) M^n >0 [M>0] If log _a M^n =z, then a^z =M^n , or , a^z =a^(xn) or, z=xn So log _a M^n =xn=n.x=nlog _a M[As log _a M=x]](https://www.tinkutara.com/question/Q221142.png)
$${Let}\:{M},{a},{n}\:{be}\:{real}\:{numbers}\:{and}\:{M}>\mathrm{0}\:,{a}>\mathrm{0},{a}\neq\mathrm{1} \\ $$$${Let}\:\mathrm{log}\underset{{a}} {\:}{M}={x} \\ $$$${So}\:{a}^{{x}} ={M}.\:\:\:\:\therefore{M}^{{n}} =\left({a}^{{x}} \right)^{{n}} ={a}^{{xn}} \\ $$$${M}^{{n}} >\mathrm{0}\:\left[{M}>\mathrm{0}\right] \\ $$$${If}\:\:\:\mathrm{log}\underset{{a}} {\:}{M}^{{n}} ={z},\:{then}\:{a}^{{z}} ={M}^{{n}} ,\: \\ $$$${or}\:,\:{a}^{{z}} ={a}^{{xn}} \\ $$$${or},\:{z}={xn} \\ $$$$\:{So}\:\mathrm{log}\underset{{a}} {\:}{M}^{{n}} ={xn}={n}.{x}={n}\mathrm{log}\underset{{a}} {\:}{M}\left[{As}\:\mathrm{log}\underset{{a}} {\:}{M}={x}\right] \\ $$