Equation logarithmic

Chipset3600

Member
Hey MHB i tried but i cant find the solution for this equation, please help me.

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

soroban

Well-known member
Hello, Chipset3600!

I think I've solved it.
Please check my work.

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

Take logs, base 5:

. . . . . . . . . . . .$$\log_5\left(27x^{\log_5x}\right) \:=\:\log_5(x^4)$$

. . . . . . .$$\log_527 + \log_5\left(x ^{\log_5x}\right) \;=\;4\:\!\log_5x$$

. . . . . . $$\log_527 + \log_5x\cdot\log_5x \;=\;4\:\!\log_5x$$

. . $$(\log_5x)^2 - 4\:\!\log_5x + \log_527 \;=\;0$$

We have a quadratic in $$\log_5x.$$

Quadratic Formula: .$$\log_5x \;=\;\frac{4 \pm \sqrt{16-4\:\!\log_527}}{2}$$

. . . . . . . . . . . . . . .$$\log_5x \;=\;2 \pm\sqrt{4-\log_527}$$

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

tkhunny

Well-known member
MHB Math Helper
Hey MHB i tried but i cant find the solution for this equation, please help me.

[TEX]{27x}^{\log_5 x}={x}^{4}[/TEX]
Note: Your coding and the appearance of the coding are not consistent. I went with the visual appearance.

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.

Chipset3600

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

SuperSonic4

Well-known member
MHB Math Helper
But, can i solve this $\sqrt{16-4\log_{5}(27)}$ without calculator?
Maybe with logarithm tables but a calculator would be much easier. Alternatively you can simplify the surd/logarithm in it's exact form
$$\sqrt{16-4\log_{5}(3^3)} = \sqrt{16-12\log_{5}(3)} = \sqrt{4(4-3\log_{5}(3))} = \sqrt{4} \times \sqrt{4-3\log_{5}(3)} = 2\sqrt{4-3\log_{5}(3)}$$

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

edit: LaTeX (ty Sudharaka and Jameson) - the error was too many of these --> }

Last edited:

Chipset3600

Member
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} 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]

So x = 3 or x=27

Opalg

MHB Oldtimer
Staff member
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
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.

Chipset3600

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