Logarithmic equation

[khavar]

x^(log(x)) = (1/100)x^3

I have been reviewing the laws of logarithms to figure out where to start. Figuring out the answer through graphing gives me the answer of x=10 or 100, yet it does not shed light on how to solve the problem algebraically.

 

[khavar]

x^3=100x^log(x)
3logx=(logx)log(100x)
3=log100x
10^3=100x
x=10

Is this work correct? How do I also find the answer x=100?

 

[JeffM]

x^(log(x)) = (1/100)x^3

I have been reviewing the laws of logarithms to figure out where to start. Figuring out the answer through graphing gives me the answer of x=10 or 100, yet it does not shed light on how to solve the problem algebraically.

Problems of the form

\(\displaystyle x^a = bx^c \implies log_d(x^a) = log_d(bx^c) \implies a * log_d(x) = log_d(b) + c * log_d(x) \implies log_d(x) * (a - c) = log_d(b).\)

Theoretically, it makes no difference what d is for that general equation, but in this particular problem a convenient d has already been selected:

\(\displaystyle x^{log_{10}(x)} = \frac{1}{100}x^3 \implies log_{10}(x) * \{log_{10}(x) - 3\} = log_{10}\left(\frac{1}{100}\right) = - 2.\)

\(\displaystyle Let\ u = log_{10}(x) \implies u(u - 3) = - 2 \implies u^2 - 3u + 2 = 0 \implies (u - 1)(u - 2) = 0 \implies u = 1\ or\ u = 2 \implies\)

\(\displaystyle log_{10}(x) = 1\ or\ log_{10}(x) = 2 \implies x = 10\ or\ x = 100.\)

Let's check.

\(\displaystyle 10^{log_{10}(10)} = 10^1 = 10^3 * 10^{-2} = \dfrac{1}{100} * 10^3.\) Looks good so far.

\(\displaystyle 100^{log_{10}(100)} = 100^2 = 10^4 = 10^6 * 10^{-2} = \dfrac{1}{100} * 100^3.\) Checks.

When you see a scary term repeated one or more times, think about whether a substitution will make things simple.

 

[mmm4444bot]

x^(log(x)) = (1/100)x^3

reviewing the laws of logarithms

it does not shed light on how to solve the problem algebraically

Take the base-10 logarithm of both sides. Use the basic laws to simplify each side of the resulting equation.

You will get a quadratic equation in log(x). Those two solutions will lead to your answers. :cool:

 

[pka]

x^(log(x)) = (1/100)x^3

Sorry, but I disagree with JeffM's notation.
Most professionals now agree that \(\displaystyle \log(x)=y\) means \(\displaystyle e^y=x\),
i.e. We no longer use the \(\displaystyle \ln(x)\) notation.

So \(\displaystyle x^{\log(x)}=\frac{x^3}{100}\) means \(\displaystyle \log(100)+[\log(x)]^2=3\log(x)\).

Can you solve \(\displaystyle u^2-3u+\log(100)=0\)

 

[soroban]

Hello, khavar!

This one requires some Olympic-level gymnastics.

\(\displaystyle \displaystyle x^{\log x} \:=\: \frac{1}{100}x^3\)

Take logs: .\(\displaystyle \log\left(x^{\log x}\right) \;=\;\log\left(\frac{1}{100}x^3\right)\)

. . . . . . . . \(\displaystyle \log x\!\cdot\!\log x \;=\;\log\left(\frac{1}{100}\right) + \log(x^3)\)

. . . . . . . . . . \(\displaystyle (\log x)^2 \;=\;\log(10^{-2}) + 3\log x\)

. . . . . . . . . . \(\displaystyle (\log x)^2 \;=\;-2 + 3\log x\)

\(\displaystyle (\log x)^2 - 3\log x + 2 \;=\;0\)

\(\displaystyle (\log x - 1)(\log x - 2) \;=\;0\)


Therefore: .\(\displaystyle \begin{Bmatrix}\log x - 1 \:=\:0 & \Rightarrow & \log x \:=\:1 & \Rightarrow & x \:=\:10\;\; \\ \log x - 2 \:=\:0 & \Rightarrow & \log x \:=\:2 & \Rightarrow & x \:=\:100 \end{Bmatrix} \)

 

[khavar]

Aha! Thank you, Jeff and M4bot. I see now that I failed to recognize that log((1/100)x^3) = log(1/100) + log(x^3). Thank you, thank you. Still angry that i couldn't see this one.

 

[khavar]

Thank you, Soroban. I appreciate the clear notation.

 

[mmm4444bot]

[Most professionals] no longer use the \(\displaystyle \ln(x)\) notation.

Maybe if Texas Instruments were to change the key label, textbook authors would begin following suit.

 

[pka]

Maybe if Texas Instruments were to change the key label, textbook authors would begin following suit.

The adjective "texas" is misleading here.
The two great twentieth century mathematicians at UTA, RL Moore and Leonard Gillman, both advocated for that definition.
Ross Perot is out of the mainstream on this one.

 

[lookagain]

x^3=100x^log(x) **

3logx=(logx)log(100x) <---This is not true, because (100x)
is not being raised to log(x) in equation**. Equation **
means 100 multiplied by the quantity x^log(x).
See that x = 100 does not check in the equation on the immediate left.
When x = 100 is substituted, the equation becomes 6 = 8. False.

x^3 = 100x^log(x) ===>

3log(x) = log[100x^log(x)] ===>

3log(x) = log(100) + log[x^log(x)] ===>

3log(x) = 2 + [log(x)][log(x)] <--- This is one of the corrected forms.

This is satisfied by x = 10 or x = 100.





Is this work correct? No, this work (meaning the whole post) is not (completely) correct.

How do I also find the answer x=100?

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