How Do You Solve the Equation log(x) = sqrt(log(x))?

[hmm?]

Hello,

I'm having trouble recovering all the solutions for this equation; here is the work I've managed to produce thus far.

logx=\sqrt{logx}

(logx)^2=logx

10^logx^2=x <--I'm not sure of this move?

x^2=x

x^2-x=0

x(x-1)=0

x= 1, 0

I know 0 cannot work, and 10 looks likes a possible solution, but I cannot produce it--anyways, any help would be appreciated.

thanks.

 

[Tide]

Why don't you try setting y = log x in the original equation then solve for y? That will avoid some confusion.

 

[hmm?]

y=logx

x=10^(logx)^1/2

This is where I'm stuck, because I feel it would be illegal to assume

10^logx^1/2...shouldn't it be (logx)^1/2. I don't know how to simplify after this point.

 

[pizzasky]

Apply Tide's hint to this equation: (logx)^2=logx

Once you have solved for y, replace y with logx, and then solve for x.

 

[VietDao29]

y=logx

x=10^(logx)^1/2

This is where I'm stuck, because I feel it would be illegal to assume

10^logx^1/2...shouldn't it be (logx)^1/2. I don't know how to simplify after this point.

Uhmm, no:
(log x)² = log(x)
By letting y = log(x), we have:
y² = y
<=> y² - y = 0
And this is our familiar quadratic equation in y. Can you solve for y from that equation? Then from there, just solve for x.
Can you go from here? :)

 

[hmm?]

Ahh, thanks! I can't believe I couldn't think of this, haha.

(logx)^2-logx=0

logx(logx-1)=0

logx=0 , log(x-1)=0

x=1, 10
A quick two questions: is there any other way to solve this, other than the y^2-y=0, and still maintain solutions 1, 10? For (logx)^n n>1, how would you treat this in an equation--merely a^n=(logx)^n never expanding?

Thanks,
Chris

 

[VietDao29]

Ahh, thanks! I can't believe I couldn't think of this, haha.

(logx)^2-logx=0

logx(logx-1)=0

logx=0 , log(x-1)=0

x=1, 10

Yes, this is correct. Congratulations.

A quick two questions: is there any other way to solve this, other than the y^2-y=0, and still maintain solutions 1, 10?

Uhm... no.
By the way, this is the common way, and is fast enough. So, I don't think you'll need some other ways though. :)

For (logx)^n n>1, how would you treat this in an equation--merely a^n=(logx)^n never expanding?

Yes, if it helps you visualize better. Say, it would be much better if you substitute y = log(x), and all of sudden, the whole equation becomes a quadratic equation, or a cubic equation,... in y. Right?
Normally, to solve an equation, we simplify the equation first to make it look less complicated. Then we isolate x to one side of the equation. Then we solve it from there.
Of course, the equation:
x = 10 ^ {\sqrt{\log x}} just looks horible, right?

Thanks,
Chris

You're welcome. :)