Equation Logarithmic: Solve x | MHB Help

[Chipset3600]

Hey MHB i tried but i can't find the solution for this equation, please help me.

$${27x}^{\log_5 x}={x}^{4}$$

 

[soroban]

Hello, Chipset3600!

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

$${27x}^{\log_5 x}=\:\:x^4$$

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]

Hey MHB i tried but i can't find the solution for this equation, please help me.

$${27x}^{\log_5 x}={x}^{4}$$

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.

$$\log\left(27x^{\log_5 x}\right)=\log\left({x}^{4}\right)$$

$$\log(27) + {\log_5 x}\cdot\log(x)=4\cdot \log(x)$$

$$\log(27) + \frac{\log(x)}{\log(5)}\cdot\log(x)=4\cdot \log(x)$$

...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]

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

 

[SuperSonic4]

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 --> }

 

[Chipset3600]

me and my friend found the solution:

$$3^{3}x^{\log_3 x } = x^{4}$$

$$x^{\log_3 x } = \frac{x^{4}}{27} doing \log_3 x = y so 3^{y} = x$$

$$27(3^{y})^{y} = (3^{y})^{y}$$

$$3^{3}.3^{y^{2}}= 3^{4y}$$
$$3^{3+y^{2}}= 3^{4y}$$
$$3+y^{2}=3^{4y}$$
$$3+y^{2}=4y$$
$$y^{2}-4y+3=0$$
$$y_{1}=1$$ $$or$$ $$y_{2}=3 \log_3 x =1 and \log_3 x = 3$$

So x = 3 or x=27

 

[Opalg]

me and my friend found the solution:

$$3^{3}x^{\log_3 x } = x^{4}$$

$$x^{\log_3 x } = \frac{x^{4}}{27}$$ doing $$\log_3 x = y$$ so $$3^{y} = x$$

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

$$3^{3}.3^{y^{2}}= 3^{4y}$$
$$3^{3+y^{2}}= 3^{4y}$$
$$3+y^{2}=3^{4y}$$
$$3+y^{2}=4y$$
$$y^{2}-4y+3=0$$
$$y_{1}=1$$ or $$y_{2}=3$$ $$\log_3 x =1$$ and $$\log_3 x = 3$$

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]

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