Solving Logarithmic Equations: Tips and Troubleshooting | Roger's Math Help

[roger]

hi

could anyone tell me where I went wrong ?

simultaneously solve

2logbase2 y = logbase4 3 + logbase2 x

3^y = 9^x

But for the top I get y = 3 root x

and bottom I get y=3x

so what's gone wrong ?


thanks

roger

 

[dextercioby]

Use \ln

The second becomes

y\ln 3=2x\ln 3\Rightarrow y=2x (1)

while the first

2\ln y=\frac{\ln 3}{2}+\ln x (2)

Solve the simple system (1) + (2)

Daniel.

 

[roger]

do you mean log base e when you state ln ?

But my working out is as follows and please tell me what went wrong :

logbase2 y ^2 = logbase2 root3 + logbase2 x

= log base 2 xroot3 = log base2 y ^2
then remove logs and square both sides to get y = 3 root x

 

[HallsofIvy]

Sorry, but I can't find anything that's gone right.

9= 3² so 9^x= 3^2x.

The second equation is 3^y= 3^2x which gives y= 2x, not y= 3x.

The first equation is 2 log₂(y)= log₄(3)+ log₂(x)
log₂(y²)- log₂(x)= log₄(3)
log₂(y²/x)= log₄(3)

If z= log₄(3) then 3= 4^z= (2²)^z= 2^2z. Now taking log₂ of both sides, gives log₂(3)= 2z so log₄(3)= (1/2)log₂(3)= log₂(3^1/2).

Putting those together, log₂(y²/x)= log₂(3^1/2) so that y²/x= 3^1/2 or y= 3^1/4x^1/2, not y= 3 x^1/2.

That is, we have 2x= 3^1/4x^1/2.

One obvious answer to that is x= 0.

Squaring both sides, 3x²= 3^1/2x and if x is not 0,
4x= 3^1/2 so x= 3^1/2/4 is another solution.

 

[dextercioby]

Halls,there's no such thing as logarithm from 0...That "x" equal 0 doesn't satisfy the first equation.

Daniel.

 

[roger]

Sorry, but I can't find anything that's gone right.

9= 3² so 9^x= 3^2x.

The second equation is 3^y= 3^2x which gives y= 2x, not y= 3x.

The first equation is 2 log₂(y)= log₄(3)+ log₂(x)
log₂(y²)- log₂(x)= log₄(3)
log₂(y²/x)= log₄(3)

If z= log₄(3) then 3= 4^z= (2²)^z= 2^2z. Now taking log₂ of both sides, gives log₂(3)= 2z so log₄(3)= (1/2)log₂(3)= log₂(3^1/2).

Putting those together, log₂(y²/x)= log₂(3^1/2) so that y²/x= 3^1/2 or y= 3^1/4x^1/2, not y= 3 x^1/2.

That is, we have 2x= 3^1/4x^1/2.

One obvious answer to that is x= 0.

Squaring both sides, 3x²= 3^1/2x and if x is not 0,
4x= 3^1/2 so x= 3^1/2/4 is another solution.

why ?

and dextercioby, x can't be negative but can't it tend to zero to give the power of negative infinity ?

 

[dextercioby]

No,no.That is x=0...It's an exact solution.It's not acceptable,as 0 is not in the domain of logarithm (in any base).

Daniel.

 

[HallsofIvy]

dexercioby was right, I was wrong- I forgot to check my answer in the original equation. The problem did not ask about limits, it asked about values for specific x. Since log₂(0) is not defined, x= 0 cannot satisfy the equation.

x= \frac{\sqrt{3}}{4}, y= \frac{\sqrt{3}}{2} is the only solution.