Question

If 2^x=3^y=12^z then proove,xy=z(x+2y)​

Answer 1

Answer:

Now, the logarithmic solution has been thoroughly explained. I'll show a way to do this without them.

Given:[math]2^x=3^y=12^z[/math]

Required to prove:

[math]x=\dfrac{2yz }{y-z}[/math]

Simplifying,

[math]\dfrac{1}{x}=\dfrac{y-z}{2yz} [/math]

[math]\dfrac{1}{x}=\dfrac{y}{2yz}- \dfrac{z} {2yz} [/math]

[math]\dfrac{1}{x}=\dfrac{1}{2z}- \dfrac{1} {2y} [/math]

[math]\dfrac{1}{x}=1/2\cdot\Big(\dfrac{1}{z}- \dfrac{1} {y}\Big) [/math]

We can divide by x,y and z because they are not zero. While this does satisfy the given but the value of the statement we are required to prove becomes undefined.

Now, let

[math]2^x=3^y=12^z=k[/math]

Then,

[math]2^x=k \rightarrow 2=k^{\frac{1}{x}}[/math]

[math]3^y=k \rightarrow 3=k^{\frac{1} {y}}[/math]

[math]12^z=k \rightarrow 12=k^{\frac{1}{z}}[/math]

So,

[math]12=k^{\frac{1}{z}}[/math]

[math]12/3=\dfrac{k^{\frac{1}{z}}}{k^{\frac{1} {y}}}[/math]

[math]4=\dfrac{k^{\frac{1}{z}}}{k^{\frac{1} {y}}}[/math]

[math]4=k^{{\frac{1}{z}}-{\frac{1} {y}}}[/math]

[math]4^{1/2}=\Big(k^{{\frac{1}{z}}-{\frac{1} {y}}}\Big)^{\frac{1}{2}}[/math]

[math]2=\Big(k^{{\frac{1}{2z}}-{\frac{1} {2y}}}\Big)[/math]

[math]k^{\frac{1}{x}}=\Big(k^{{\frac{1}{2z}}-{\frac{1} {2y}}}\Big)[/math]

Comparing exponents,

[math]\dfrac{1}{x}=\Big(\dfrac{1}{2z}-\dfrac{1} {2y}\Big)[/math]

[Remember this?]

[math]\dfrac{1}{x}=\dfrac{y-z}{2yz}[/math]

[math]x=\dfrac{2yz}{y-z}[/math]