Question

solve the simultaneous equation log2(x)+log2(y)=4 log2(x)+2log2(y)=3​

Answer 1

Question:

Solve the simultaneous equations:

$\displaystyle{\sf\:\log_2\:(\:x\:)\:+\:\log_2\:(\:y\:)\:=\:4}$

$\displaystyle{\sf\:\log _2\:(\:x\:)\:+\:2\:\log_2\:(\:y\:)\:=\:3}$

Answer:

The solution of the given simultaneous equations is

$\displaystyle{\boxed{\red{\sf\:(\:x\:,\:y\:)\:=\:\left(\:32\:,\:\dfrac{1}{2}\:\right)\:}}}$

Step-by-step-explanation:

We have given logarithmic simultaneous equations.

The given simultaneous equations are

$\displaystyle{\sf\:\log_2\:(\:x\:)\:+\:\log_2\:(\:y\:)\:=\:4\:\quad\cdots(\:1\:)}$

$\displaystyle{\sf\:\log _2\:(\:x\:)\:+\:2\:\log_2\:(\:y\:)\:=\:3\:\quad\cdots(\:2\:)}$

Now,

$\displaystyle{\sf\:\log_2\:(\:x\:)\:+\:\log_2\:(\:y\:)\:=\:4\:\quad\cdots(\:1\:)}$

$\displaystyle{\implies\sf\:log_2\:(\:xy\:)\:=\:4}$

We know that,

$\displaystyle{\sf\:2^4\:=\:16}$

$\displaystyle{\implies\sf\:\log_2\:(\:2^4\:)\:=\:\log_2\:(\:16\:)}$

$\displaystyle{\implies\sf\:4\:\log_2\:(\:2\:)\:=\:\log_2\:(\:16\:)\:\qquad\cdots\:[\:\log_a\:(\:b^k\:)\:=\:k\:\log_a\:(\:b\:)\:]}$

$\displaystyle{\implies\sf\:4\:=\:\log_2\:(\:16\:)}$

By using this value, we get,

$\displaystyle{\sf\:log_2\:(\:xy\:)\:=\:4}$

$\displaystyle{\implies\sf\:\log_2\:(\:xy\:)\:=\:\log_2\:(\:16\:)}$

By taking antilog to the base 2 on both sides,

$\displaystyle{\implies\sf\:xy\:=\:16}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{16}{y}\:\qquad\cdots(\:3\:)}$

Now,

$\displaystyle{\sf\:\log _2\:(\:x\:)\:+\:2\:\log_2\:(\:y\:)\:=\:3\:\quad\cdots(\:2\:)}$

$\displaystyle{\implies\sf\:\log_2\:(\:x\:)\:+\:\log_2\:(\:y^2\:)\:=\:3\:\qquad\cdots\:[\:k\:\log_a\:(\:b\:)\:=\:\log_a\:(\:b^k\:)\:]}$

$\displaystyle{\implies\sf\:\log_2\:(\:xy^2\:)\:=\:3}$

We know that,

$\displaystyle{\sf\:2^3\:=\:8}$

$\displaystyle{\implies\sf\:\log_2\:(\:2^3\:)\:=\:\log_2\:(\:8\:)}$

$\displaystyle{\implies\sf\:3\:\log_2\:(\:2\:)\:=\:\log_2\:(\:8\:)\:\qquad\cdots\:[\:\log_a\:(\:b^k\:)\:=\:k\:\log_a\:(\:b\:)\:]}$

$\displaystyle{\implies\sf\:3\:=\:\log_2\:(\:8\:)}$

By using this value, we get,

$\displaystyle{\sf\:log_2\:(\:xy^2\:)\:=\:3}$

$\displaystyle{\implies\sf\:\log_2\:(\:xy^2\:)\:=\:\log_2\:(\:8\:)}$

By taking antilog to the base 2 on both sides,

$\displaystyle{\implies\sf\:xy^2\:=\:8}$

$\displaystyle{\implies\sf\:\dfrac{16}{y}\:\times\:y^2\:=\:8\:\qquad\cdots[\:From\:(\:3\:)\:]}$

$\displaystyle{\implies\sf\:\dfrac{16}{\cancel{y}}\:\times\:\cancel{y}\:\times\:y\:=\:8}$

$\displaystyle{\implies\sf\:16y\:=\:8}$

$\displaystyle{\implies\sf\:y\:=\:\cancel{\dfrac{8}{16}}}$

$\displaystyle{\implies\boxed{\pink{\sf\:y\:=\:\dfrac{1}{2}\:}}}$

By substituting this value in equation ( 3 ), we get,

$\displaystyle{\implies\sf\:x\:=\:\dfrac{16}{y}\:\qquad\cdots(\:3\:)}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{16}{\dfrac{1}{2}}}$

$\displaystyle{\implies\sf\:x\:=\:16\:\times\:2}$

$\displaystyle{\implies\boxed{\blue{\sf\:x\:=\:32\:}}}$

∴ The solution of the given simultaneous equations is

$\displaystyle{\underline{\boxed{\red{\sf\:(\:x\:,\:y\:)\:=\:\left(\:32\:,\:\dfrac{1}{2}\:\right)\:}}}}$