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#### Chipset3600

##### Member

- Feb 14, 2012

- 79

[TEX]{27x}^{\log_5 x}={x}^{4}[/TEX]

- Thread starter Chipset3600
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- Thread starter
- #1

- Feb 14, 2012

- 79

[TEX]{27x}^{\log_5 x}={x}^{4}[/TEX]

I think I've solved it.

Please check my work.

[TEX]{27x}^{\log_5 x}=\:\:x^4[/TEX]

Take logs, base 5:

. . . . . . . . . . . .[tex]\log_5\left(27x^{\log_5x}\right) \:=\:\log_5(x^4)[/tex]

. . . . . . .[tex]\log_527 + \log_5\left(x ^{\log_5x}\right) \;=\;4\:\!\log_5x[/tex]

. . . . . . [tex]\log_527 + \log_5x\cdot\log_5x \;=\;4\:\!\log_5x[/tex]

. . [tex](\log_5x)^2 - 4\:\!\log_5x + \log_527 \;=\;0 [/tex]

We have a quadratic in [tex]\log_5x.[/tex]

Quadratic Formula: .[tex]\log_5x \;=\;\frac{4 \pm \sqrt{16-4\:\!\log_527}}{2} [/tex]

. . . . . . . . . . . . . . .[tex]\log_5x \;=\;2 \pm\sqrt{4-\log_527}[/tex]

Therefore: .[tex]x \;=\;5^{2\pm\sqrt{4-\log_527}} \;=\; \begin{Bmatrix}236.8886726 \\ 2.638370139 \end{Bmatrix} [/tex]

Note: Your coding and the appearance of the coding are not consistent. I went with the visual appearance.

[TEX]{27x}^{\log_5 x}={x}^{4}[/TEX]

A few ways to go about it. I kind of liked introducing another logarithm. There's already so much, that this may not come to mind.

[TEX]\log\left(27x^{\log_5 x}\right)=\log\left({x}^{4}\right)[/TEX]

[TEX]\log(27) + {\log_5 x}\cdot\log(x)=4\cdot \log(x)[/TEX]

[TEX]\log(27) + \frac{\log(x)}{\log(5)}\cdot\log(x)=4\cdot \log(x)[/TEX]

...and it's magically Quadratic in log(x). You should be able to solve that.

Let's see what you get unless someone else does all the work for you.

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- #4

- Feb 14, 2012

- 79

But, can i solve this [TEX]\sqrt{16-4\log_{5}(27)}}[\TEX] without calculator?

- Mar 1, 2012

- 249

Maybe with logarithm tables but a calculator would be much easier. Alternatively you can simplify the surd/logarithm in it's exact formBut, can i solve this $\sqrt{16-4\log_{5}(27)}$ without calculator?

$$ \sqrt{16-4\log_{5}(3^3)} = \sqrt{16-12\log_{5}(3)} = \sqrt{4(4-3\log_{5}(3))} = \sqrt{4} \times \sqrt{4-3\log_{5}(3)} = 2\sqrt{4-3\log_{5}(3)} $$

Leaving it in exact form is usually better than rounding it off to a decimal.

edit: LaTeX (ty

Last edited:

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- #6

- Feb 14, 2012

- 79

[TEX]3^{3}x^{\log_3 x } = x^{4}[/TEX]

[TEX]x^{\log_3 x } = \frac{x^{4}}{27} doing \log_3 x = y so 3^{y} = x[/TEX]

[TEX]27(3^{y})^{y} = (3^{y})^{y}[/TEX]

[TEX]3^{3}.3^{y^{2}}= 3^{4y}[/TEX]

[TEX]3^{3+y^{2}}= 3^{4y}[/TEX]

[TEX]3+y^{2}=3^{4y}[/TEX]

[TEX]3+y^{2}=4y[/TEX]

[TEX]y^{2}-4y+3=0[/TEX]

[TEX]y_{1}=1[/TEX] [TEX]or[/TEX] [TEX]y_{2}=3 \log_3 x =1 and \log_3 x = 3[/TEX]

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- #7

- Feb 7, 2012

- 2,792

Good solution (apart from the typo). But you misled everyone here by stating the problem wrongly. You wrote ${27x}^{\log_{\color{red}5} x}={x}^{4}$, and the 5 should have been 3.me and my friend found the solution:

[TEX]3^{3}x^{\log_3 x } = x^{4}[/TEX]

[TEX]x^{\log_3 x } = \frac{x^{4}}{27}[/TEX] doing [TEX]\log_3 x = y[/TEX] so [TEX]3^{y} = x[/TEX]

[TEX]27(3^{y})^{y} = (3^{y})^{y}[/TEX] That last $\color{red}y$ should be a 4.

[TEX]3^{3}.3^{y^{2}}= 3^{4y}[/TEX]

[TEX]3^{3+y^{2}}= 3^{4y}[/TEX]

[TEX]3+y^{2}=3^{4y}[/TEX]

[TEX]3+y^{2}=4y[/TEX]

[TEX]y^{2}-4y+3=0[/TEX]

[TEX]y_{1}=1[/TEX] or [TEX]y_{2}=3[/TEX] [TEX]\log_3 x =1[/TEX] and [TEX]\log_3 x = 3[/TEX]

So x = 3 or x=27

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- #8

- Feb 14, 2012

- 79

OMG! Sorry guys! it was my mistake, the base of the Log is 3 and not 5...Thanks