ขอช่วยหน่อยครับ

[pogpagasd]

ตามนี้ครับ

[lnพwsะบุ๑sสุ๑xล่o]

1.

$xyz=10^{81}$

$logxyz=81$

$logx+logy+logz=81$



$(log_{10}x)(log_{10}yz)+(log_{10}y)(log_{10}z)=468$

$(log_{10}x)(log_{10}y+log_{10}z)+(log_{10}y)(log_{10}z)=468$

$logxlogy+logylogz+logzlogx=468$



$\sqrt{(log_{10}x)^2+(log_{10}y)^2+(log_{10}z)^2} $

$= \sqrt{(logx+logy+logz)^2-2(logxlogy+logylogz+logzlogx)} $

$= \sqrt{81^2-2(468)}=75$

[lnพwsะบุ๑sสุ๑xล่o]

2.

Take log to all equations;

$(log_37)(loga)=log27=3log3$

$\frac{log7}{log3}(loga)=3log3$

$loga=\frac{3(log3)^2}{log7}$

$a=10^{\frac{3(log3)^2}{log7}}$

In the same rhythm , we get

$(log_711)(logb)=log49=2log7$

$b=10^{\frac{2(log7)^2}{log11}}$

$(log_{11}25)(logc)=0.5(log11)$

$c=10^{\frac{0.5(log11)^2}{log25}}$


$a^{(log_37)^{2}}+b^{(log_{7}11)^{2}}+c^{(log_{11}{25})^{2}}$

$=10^{3log7}+10^{2log11}+10^{0.5log25}$

$=7^3+11^2+25^{0.5}=343+121+5=469$

[lnพwsะบุ๑sสุ๑xล่o]

3.

ให้ $log_{225}x=a$ และ $log_{64}y=b$

จะได้

$a+b=4...(1)$

$\frac{1}{a} -\frac{1}{b} =1...(2)$

แทน $a=4-b$ ใน (2)

$\frac{1}{4-b} -\frac{1}{b} =1$

$2b-4=4b-b^2$

$b^2-2b-4=0$

$(b-1)^2=5$

$b-1=\pm \sqrt{5} $

$b=1\pm \sqrt{5} $

$a=3\mp \sqrt{5} $

$log_{30}(x_1y_1x_2y_2)=log_{30}(225^{a_1+a_2}+64^{b_1+b_2})$

$=log_{30}(225^6+64^2)$

[pogpagasd]

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