Solving A Homemade Exponential Equation

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10^{log(x)}^2=x
#ChallengingMathProblems #ExponentialEquations #Logarithms
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[muhammademran]

Here best method:

10^(logx)^2 = x
Taking log

Log x = (logx)^2

Log x. Logx - log x = 0

So log x =0 or log x =1

X = 10^0 = 1 or x = 10^1 = 10

[ManjulaMathew-wbzn]

One last method 3A Since x can be written as 10^log x the original equation becomes
10^((logx)^2)=10^(logx) now comparing exponents
(logx)^2=logx and the rest is routine.

[RexxSchneider]

Before watching: take log of both sides, assuming log is base 10.
(logx)^2 = logx. So (logx)^2 - logx = 0. Therefore logx * (logx - 1) = 0.
Either logx = 0 or logx = 1. So x = 1 or x = 10.
After watching: method 1 and method 2 are actually equivalent and only require mental arithmetic. Method 3 is unusual, but interesting.

[sngmn]

10^lg^2(x) = 10^(lg(x) * lg(x)) = x ^ lg(x) => x ^ lg(x) = x

Lg both sides:

lg^2(x) = lg(x)

Probably not only me, but also others really want to reduce this by a logarithm))) OK. But, before that we must know that we cant divide by zero)) So go and check what happens when lg(x) = 0? lg(x) = 0 => x = 1 and it is a solution)).
OK, now suppose lg(x) != 0 and bravely divide both sides by lg(x): lg(x) = 1 => x = 10. We got second one.

[mcwulf]

Usually I like to avoid using logs or lns in these types of problem but this time I did your method 2. Your method 3 is neat.

[marklevin]

Hocus pocus transforms our equation into x^2-11x+10=0 Simple quadratic formula gives us roots 1 and 10

[benreal]

For this problem, let's just analyze 10^log(x) = x first. Because log has a base of 10, 10^log(x) just cancels out to be x. Hence x = x. This is an identity. However, here we're applying log(x)^2. Because we have an identity already, the value of the log(x) cannot change. Therefore, the log(x) can equal 0 or 1, as 1^2 = 1, and 0^2 = 0. Any other value will result in a change. log(x) = 1 -> x = 10. log(x) = 0 -> x = 1. Therefore x=10 and x=1 are our solutions.

[robertlunderwood]

Just as a CYA move on the third method, check for plausible solutions with 0 as a base to see if a non zero exponent is possible or with -1 as a base with an even exponent. There are no solutions with these possibilities in this problem, but I've been bit more than once.

[StaR-uwdc]

Fourth method:
x=10^logx
10^(logx)^2 = 10^logx
(logx)^2=logx
logx(logx-1)=0
logx = 0 or 1 so x=1 or 10

[Paul-]

8:11 x^(log x) is not equal to x in general. You just got lucky because your two solutions (1, 10) happen to be the two values where that is true.

[danhgiasanpham]

Thank you for sharing. Let's stay connected for long-term interaction.

[capitano]

I've transformed the logx into lnx/ln10 and did ln ()on porch sides to get out the powerlog

[Anmol_Sinha]

I used the third method to solve this and now feel stupid for not using the 2nd one lol.

[Packerfan]

10^(log x)^2 = x
log 10^(log x)^2 = log x
(log x)^2 log 10 = log x
(log x)^2 = log x
(log x)^2 - log x = 0
(log x)(log x - 1) =0
log x = 0 or log x = 1
x = 1 or x = 10