Logorythm equations

**dirtyharry** · December 12th 2012, 08:49 AM

$lgx^{4} - \frac{30}{lgx} = 2$

I tried making lgx = y, but I can't get it to work.

Next problem is:

$log_{2}x - log_{4}x + log_{16}x = 3$

I managed to get it down to $\frac{4}{4log_{x^{2}}} - \frac{2}{4log_{x^{2}}} + \frac{1}{4log_{x^{2}}} = 3$

but I don't know what next. The end result should be 16.

And what about these two:

$\frac{z}{2 + lgx} + \frac{z}{4-lgx} = 1$

and

$log_{x}3 + log_{3}x = 2$

What I did, was $log_{x}3 + \frac{1}{log_{x}3} = 2$

Then, $log_{x}3 = y$ and that's where I'm stuck. What I get is: $y+\frac{1}{y}=2$ I can't get a squared y out of it. And I don't know how to deal with the fraction.
Please reply,
Thanks!

**skeeter** · December 12th 2012, 10:43 AM

$\log{x^4} - \frac{30}{\log{x}} = 2$

$4\log{x} - \frac{30}{\log{x}} = 2$

$4\log^2{x} - 30 = 2\log{x}$

$2\log^2{x} - \log{x} - 15 = 0$

$(2\log{x} + 5)(\log{x} - 3) = 0$

finish it ...

$\log_2{x} - \log_4{x} + \log_{16}{x} = 3$

use a change of base ...

$\log_2{x} - \frac{\log_2{x}}{\log_2{4}} + \frac{\log_2{x}}{\log_2{16}} = 3$

$\log_2{x} - \frac{\log_2{x}}{2} + \frac{\log_2{x}}{4} = 3$

finish it ...

**Soroban** · December 12th 2012, 10:44 AM

Hello, dirtyharry!

Here is the second one . . .

$\log_2x - \log_4x + \log_{16}x \:=\: 3$

Base-change Formula:

n $\frac{\log_2x}{\log_22} - \frac{\log_2x}{\log_24} + \frac{\log_2x}{\log_216} \;=\;3$

. . $\frac{\log_2x}{1} - \frac{\log_2x}{2} + \frac{\log_2x}{4} \;=\;3$

. . . . . . . . . . . . . $\tfrac{3}{4}\log_2x \;=\;3$

. . . . . . . . . . . . . . $\log_2x \;=\;4$

. . . . . . . . . . . . . . . . . $x \;=\;2^4 \;=\;16$

**earboth** · December 12th 2012, 10:52 AM

Originally Posted by dirtyharry

...

$log_{x}3 + log_{3}x = 2$

What I did, was $log_{x}3 + \frac{1}{log_{x}3} = 2$

Then, $log_{x}3 = y$ and that's where I'm stuck. What I get is: $y+\frac{1}{y}=2$ ...

Here is the last one:

All your calculations are correct!

$y+\frac{1}{y}=2$ ................. multiply through by y:

$y^2-2y+1=0~\implies~(y-1)^2=0$

The soultion of this equation should be obvious.

Now solve for x:

$log_{x}3 = 1$

**earboth** · December 12th 2012, 11:19 AM

Originally Posted by dirtyharry

...

And what about these two:

$\frac{z}{2 + lgx} + \frac{z}{4-lgx} = 1$

...

This one is the best!

I assume that $\lg(x) = \log_{10}(x)$

$\frac{z(4-\lg(x))}{(2 + \lg(x))(4-\lg(x))} + \frac{z(2+\lg(x))}{(4-\lg(x))(2+\lg(x))} = 1$

Multiply through by the common denominator, expand the brackets and collect like terms at the LHS. You should come out with:

$(\lg(x))^2-2 \lg(x) + 6z -8 = 0$ ............. Replace lg(x) by y:

$y^2-2y+6z-8=0$ ............ Solve for y:

$y = 1 \pm \sqrt{9-6z}$

Therefore: $z \le \frac32$

If

$z = \frac32~\implies~y = 1~\implies~x = 10$

$z = \frac43~\implies~y = 2 \ \vee \ y=0 ~\implies~ x = 100 \ \vee \ x=1$

$z = 0~\implies~y = 4 \ \vee \ y=-2 ~\implies~ x = 10000 \ \vee \ x=\frac1{100}$

etc.

**dirtyharry** · December 12th 2012, 02:51 PM

Much appreciated

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