Solving Logarithmic Equations: How to Find the Value of x

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Homework Statement



logx+log(x+3)=1

Homework Equations





The Attempt at a Solution



Ive tried to put it in solve so i know x=2

i think i have to, hmm i don't know if its called differantiate in english but i don't know how to differantiate log.


Edit: this is what I've tried after differantiating

1/x+1/x=1
that means it must be 2 since 2 halfs give 1
 
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Welcome to Physics Forums.

I'm guessing that you're trying to solve for equation to find x?
 


yes, sry for my bad explaining of my problem
 


HalloGubi said:
yes, sry for my bad explaining of my problem
No problem.

When attempting to solve equations, usually the best thing to do is to collect all the terms you want together. So can you combine the two log terms into a single logarithm?
 


HalloGubi said:

Homework Statement



logx+log(x+3)=1

Homework Equations





The Attempt at a Solution



Ive tried to put it in solve so i know x=2

i think i have to, hmm i don't know if its called differantiate in english but i don't know how to differantiate log.


Edit: this is what I've tried after differantiating

1/x+1/x=1
that means it must be 2 since 2 halfs give 1

This problem has nothing to do with differentiation. All you need to know are the properties of logarithms and how to factor quadratics.
 


I know this is an old post but incase people find it in the future:

rule 1)
logbx + logby = logbxy

rule 2)
x (x + a) = x2 + ax

rule 3)
solving quadratic equations
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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