Algebra (Olympiad)hardFree

Logarithmic System of Equations: Evaluate y1 + y2 | IOQM

JEE Maths question with a full step-by-step solution.

Question
The system of equations
log⁡10(2000xy)−(log⁡10x)(log⁡10y)=4,\log_{10}(2000xy)-(\log_{10}x)(\log_{10}y)=4,
log⁡10(2yz)−(log⁡10y)(log⁡10z)=1,\log_{10}(2yz)-(\log_{10}y)(\log_{10}z)=1,
log⁡10(zx)−(log⁡10z)(log⁡10x)=0\log_{10}(zx)-(\log_{10}z)(\log_{10}x)=0
 has two solutions in real numbers (x1,y1,z1)(x_1,y_1,z_1) and (x2,y2,z2)(x_2,y_2,z_2). Evaluate y1+y2y_1+y_2.
Solution
Answer: 25
Step 1: Substitute a=log⁡10x, b=log⁡10y, c=log⁡10za=\log_{10}x,\ b=\log_{10}y,\ c=\log_{10}z, and expand each logarithm of a product. Since log⁡102000=3+log⁡102\log_{10}2000=3+\log_{10}2,
3+log⁡102+a+b−ab=4 ⇒ log⁡102+a+b−ab=1....(1)3+\log_{10}2+a+b-ab=4\ \Rightarrow\ \log_{10}2+a+b-ab=1.\qquad\text{...(1)}
 Similarly, from log⁡10(2yz)=log⁡102+b+c\log_{10}(2yz)=\log_{10}2+b+c,
log⁡102+b+c−bc=1....(2)\log_{10}2+b+c-bc=1.\qquad\text{...(2)}
 And from log⁡10(zx)=c+a\log_{10}(zx)=c+a,
c+a−ca=0....(3)c+a-ca=0.\qquad\text{...(3)}
 Step 2: Solve (3) for cc. From c(1−a)=−ac(1-a)=-a,
c=aa−1.c=\frac{a}{a-1}.
 Step 3: Rewrite (2) using 1−log⁡102=log⁡1051-\log_{10}2=\log_{10}5, so b+c−bc=log⁡105b+c-bc=\log_{10}5. Solving for bb,
b=log⁡105−c1−c.b=\frac{\log_{10}5-c}{1-c}.
 Substituting c=aa−1c=\dfrac{a}{a-1} (so that 1−c=−1a−11-c=-\dfrac{1}{a-1}),
b=a+(1−a)log⁡105....(4)b=a+(1-a)\log_{10}5.\qquad\text{...(4)}
 Step 4: Put (4) into (1). Since log⁡102+a+b−ab=1\log_{10}2+a+b-ab=1 gives a+b(1−a)=1−log⁡102=log⁡105a+b(1-a)=1-\log_{10}2=\log_{10}5, we have
a+(1−a)[a+(1−a)log⁡105]=log⁡105,a+(1-a)\big[a+(1-a)\log_{10}5\big]=\log_{10}5,
 which expands to
2a−a2+(1−a)2log⁡105=log⁡105.2a-a^2+(1-a)^2\log_{10}5=\log_{10}5.
 Bringing the log⁡105\log_{10}5 terms together,
(2a−a2)log⁡105=2a−a2 ⇒ (2a−a2)(log⁡105−1)=0.(2a-a^2)\log_{10}5=2a-a^2\ \Rightarrow\ (2a-a^2)(\log_{10}5-1)=0.
 Step 5: Since log⁡105≠1\log_{10}5\neq1, we need 2a−a2=02a-a^2=0, i.e. a(2−a)=0a(2-a)=0, so a=0a=0 or a=2a=2. Step 6: Find bb in each case from (4):
a=0: b=0+(1)log⁡105=log⁡105,a=2: b=2+(−1)log⁡105=2−log⁡105.a=0:\ b=0+(1)\log_{10}5=\log_{10}5,\qquad a=2:\ b=2+(-1)\log_{10}5=2-\log_{10}5.
 Step 7: Recover y=10 by=10^{\,b}:
b=log⁡105 ⇒ y=5,b=2−log⁡105 ⇒ y=1025=20.b=\log_{10}5\ \Rightarrow\ y=5,\qquad b=2-\log_{10}5\ \Rightarrow\ y=\frac{10^2}{5}=20.
 Therefore y=5y=5 or y=20y=20, and
y1+y2=5+20=25.y_1+y_2=5+20=25.
 Answer: 25
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