# Equation Logarithmic: Solve x | MHB Help

• MHB
• Chipset3600
In summary, the equation given in the conversation is {27x}^{\log_3 x}={x}^{4}, and using the quadratic formula, the solutions are x=3 or x=27. This was found by first taking logs, base 3, and simplifying the expression to a quadratic in \log_3 x. The solutions were then obtained by solving for \log_3 x using the quadratic formula and substituting back in for x.
Chipset3600

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

Hello, Chipset3600!

I think I've solved it.

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

Chipset3600 said:

[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.

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

Chipset3600 said:
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:
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

Chipset3600 said:
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.

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

## 1. What is a logarithmic equation?

A logarithmic equation is an equation in which the variable appears as the exponent of a logarithm. It can be written in the form logb(x) = y, where b is the base of the logarithm and x and y are the inputs and outputs of the equation.

## 2. How do you solve a logarithmic equation?

To solve a logarithmic equation, you first need to isolate the logarithm on one side of the equation. Then, you can rewrite the logarithm in exponential form, using the base as the exponent. Finally, solve for the variable by using inverse operations.

## 3. What is the difference between a logarithmic and exponential equation?

The main difference between logarithmic and exponential equations is the placement of the variable. In a logarithmic equation, the variable is the exponent, while in an exponential equation, the variable is the base.

## 4. Can logarithmic equations have more than one solution?

Yes, logarithmic equations can have more than one solution. This is because the logarithm function is one-to-one, meaning that different inputs can result in the same output.

## 5. What is the significance of the base in a logarithmic equation?

The base in a logarithmic equation determines the relationship between the input and output values. For example, a base of 2 means that the input is being multiplied by 2 each time, while a base of 10 means that the input is being multiplied by 10 each time.

• General Math
Replies
3
Views
1K
• General Math
Replies
1
Views
920
• Feedback and Announcements
Replies
42
Views
4K
• General Math
Replies
2
Views
10K
• General Math
Replies
1
Views
837
• General Math
Replies
2
Views
1K
• General Math
Replies
1
Views
1K
• General Math
Replies
1
Views
5K
• General Math
Replies
20
Views
2K
• General Math
Replies
1
Views
1K