Berikut ini adalah kumpulan soal mengenai persamaan logaritma tipe HOTS. Jika ingin bertanya soal, silahkan gabung ke grup
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No. 1
Misalkan
x ,
y ,
z bilangan real positif yang memenuhi sistem persamaan:
3\ {^x\negmedspace\log\left(3y\right)}=3\left(^{3x}\negmedspace\log(27z)\right)={^{3x^4}\negmedspace\log(81yz)}\ne0
Nilai
x^5y^4z adalah ....
Alternatif Penyelesaian
3\ {^x\negmedspace\log\left(3y\right)}=3\left(^{3x}\negmedspace\log(27z)\right)={^{3x^4}\negmedspace\log(81yz)}=k
\(\eqalign{
3\ {^x\negmedspace\log\left(3y\right)}&=k\\
{^x\negmedspace\log\left(3y\right)}&=\dfrac{k}3\\
3y&=x^{\frac{k}3}
}\)
\(\eqalign{
3\left(^{3x}\negmedspace\log(27z)\right)&=k\\
^{3x}\negmedspace\log(27z)&=\dfrac{k}3\\
27z&=(3x)^{\frac{k}3}\\
&=3^{\frac{k}3}x^{\frac{k}3}
}\)
\(\eqalign{
{^{3x^4}\negmedspace\log(81yz)}&=k\\
81yz&=\left(3x^4\right)^k\\
3y\cdot27z&=3^kx^{4k}\\
x^{\frac{k}3}\cdot3^{\frac{k}3}x^{\frac{k}3}&=3^kx^{4k}\\
3^{\frac{k}3}x^{\frac{2k}3}&=3^kx^{4k}\\
\left(3^{\frac{k}3}x^{\frac{2k}3}\right)^{\frac3k}&=\left(3^kx^{4k}\right)^{\frac3k}\\
3x^2&=3^3x^{12}\\
x^{10}&=3^{-2}\\
x&=3^{-\frac15}
}\)
\(\eqalign{
3\ {^x\negmedspace\log\left(3y\right)}&={^{3x^4}\negmedspace\log(81yz)}\\
{^{3^{-\frac15}}\negmedspace\log\left(3y\right)^3}&={^{3\left(3^{-\frac15}\right)^4}\negmedspace\log\left(3^4yz\right)}\\
-5{^3\negmedspace\log\left(3^3y^3\right)}&={^{3\left(3^{-\frac45}\right)}\negmedspace\log\left(3^4yz\right)}\\
-5{^3\negmedspace\log\left(3^3y^3\right)}&={^{3^{\frac15}}\negmedspace\log\left(3^4yz\right)}\\
-5{^3\negmedspace\log\left(3^3y^3\right)}&=5\ {^3\negmedspace\log\left(3^4yz\right)}\\
-{^3\negmedspace\log\left(3^3y^3\right)}&={^3\negmedspace\log\left(3^4yz\right)}\\
3^{-3}y^{-3}&=3^4yz\\
3^{-7}&=y^4z
}\)
\(\eqalign{
x^5y^4z&=\left(3^{-\frac15}\right)^53^{-7}\\
&=3^{-1}3^{-7}\\
&=3^{-8}\\
&=\dfrac1{3^8}\\
&=\boxed{\boxed{\dfrac1{6561}}}
}\)
1 Komentar
Luar biasa mas
BalasHapusSilahkan berkomentar dengan santun di sini. Anda juga boleh bertanya soal matematika atau mengoreksi jawaban di atas