The function $f(x)=4^x+6^x-9^x$ is such that $f(0)=1&gt;0, f(1)=1&gt;0, f(2)=-29$ and next $g(x)=(4/9)^x+(6/9)^x-1 \implies f'(x)&lt;0$ for all real values of $x$. So $g(x)$ being monotonic the equation $$4^x+6^x=9^x$$ has exactly one real solution. The question is whether this real root can be found analytically by hand.
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So here is the problem      Solve $$4^x + 6^x = 9^x$$ for $x$.   I am trying to find its real solution.  I was trying in this way!! $$6^x\left( \frac {4^x}{6^x} + 1\right) = 9^x$$ $$\left({\frac 23}\right)^x +1=\left({\frac 32}\right)^x$$ but I'm stuck . . . Will anyone help ?
82616
This equation  $$3^x+4^x=2^x+5^x$$ has two obvious real roots. The question is if it has more real roots than two. A proof is required in any case.
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How to get $f(x)$, if we know that $f(f(x))=x^2+x$?  Is there an elementary function $f(x)$ that satisfies the equation?
92723
$$f(x)= 1 - e^x\sin(x)$$   $$g(x)= 1 + e^x\cos(x)$$   Prove that between any two roots of $f$ there exists at least one root of $g$.   I know Rolle's Theorem and the Intermediate Value Theorem (I think) need to be used. Can someone show a step by step proof for this?
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$P(x) = 0$ is a polynomial equation having at least one integer root, where $P(x)$ is a polynomial of degree five and having integer coefficients.  If $P(2) = 3$ and $P(10)= 11$, then prove that the equation $P(x) = 0$ has exactly one integer root.  I tried by assuming a fifth degree polynomial but got stuck after that.  The question was asked by my friend.
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Let  $f(xy) =f(x)f(y)$ for all $x,y\geq 0$. Show that $f(x) = x^p$ for some $p$.    I am not very experienced with proof. If we let $g(x)=\log (f(x))$ then this is the same as  $g(xy) = g(x) + g(y)$   I looked up the hint and it says let $g(x) = \log f(a^x) $   The wikipedia page for functional equations only states the form of the solutions without proof.  Attempt  Using the hint (which was like pulling a rabbit out of the hat)  Restricting the codomain $f:(0,+\infty)\rightarrow (0,+\infty)$ so that we can define the real function $g(x) =  \log f(a^x)$ and we have $$g(x+y) = g(x)+ g(y)$$   i.e $g(x) = xg(1)$ as $g(x)$ is continuous (assuming $f$ is).  Letting $\log_a f(a) = p$ we get $f(a^x) =a^p $. I do not have a rigorous argument but I think I can conclude that $f(x) = x^p$ (please fill any holes or unspecified assumptions) Different solutions are invited
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Lets say I have a function defined as $f(x) = \sqrt {x^2}$. Common knowledge of square roots tells you to simplify to $f(x) = x$ (we'll call that $g(x)$) which may be the same problem, but it isn't the same equation. For example, say I put $-1$ into them:  $\begin{align} f(x) &amp;= \sqrt {x^2} \\ f(-1) &amp;= \sqrt {(-1)^2} \\ f(-1) &amp;= \sqrt {1} \\ f(-1) &amp;= 1 \end{align}$  $\begin{align} g(x) &amp;= x \\ g(-1) &amp;= -1 \end{align}$  thereby, we conclude that $f(x)$ and $g(x)$ do not produce the same results even though they are mathematically the same. This is also shown when we try to graph the functions:  :    :    :    From this, we can see that given $f(x) = \sqrt {x^2}$, when simplified is not the same as $f(x) = x$. So, is there any way to prove that $y = \sqrt {x^2}$ is not the same as $y = x$ for negative values, but is infact the same as $y = \mid x \mid$?
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Prove that given $ a &lt; b &lt; c $ this equation:  $$\frac{1}{x-a} + \frac{1}{x-b} + \frac{1}{x-c} = 0 $$   has precisely 2 real roots.   I understand there are 3 point of discontinuities, but I have no idea how to prove this. Can you give me a hint?   Thanks in advance.
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Consider the equation $$|2^x-3^y|=1$$ in the unknowns $x \in \mathbb{N}$ and $y \in \mathbb{N}$. Is it possible to prove that the only solutions are $(1,1)$, $(2,1)$ and $(3,2)$?
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Is there any "superlogarithm" or something to solve an equation like this:  $$x^x = 10?$$
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Problem is as stated in the title. Source is Larson's 'Problem Solving through Problems'. I've tried all kinds of factorizations with this trying to get it to the form $$k^2l^2$$ but nothing's clicking. I tried Bézout but the same expression can be written as $$a^2 + (a^2 + 1)(a+1)^2$$ which would imply that there is no real root.  Would really appreciate some help, thanks.
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How can I show that the equation  $$e^x-\ln(x)-2^{2014}=0$$ has exactly two positive roots?
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A certain physics problem I have been working on has turned into a math problem. Particularly, I want to find the solutions of some equation of the form   $$f(x)=ae^{bx}+cx+d = 0$$  where $a, b, c,$ and $d$ are constant, real numbers that come from the physics problem and $x$ will be a real number. I do not know how to find the solutions of an equation of this form. In the special case where $c=0$, the solution is simply  $$x = \frac{\ln(\frac{-d}{a})}{b}$$  but for other values of $c$, I am stumped. If there is no exact form for the roots, is there a relatively simple way to estimate the roots?
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Based on pure intuition, the root is 0. It’s very obvious. But what is the proper way to solve it? Also how do we know it’s the only possible root (the answer is 1 root only, but how can I confirm?)
47757
Im trying to solve a problem X.12.6 from Sarason's Complex Function Theory:  So I tried using Rouché's theorm, on the function $g(z)=z^n$, on the domain $\{z:z&lt;\sqrt{1+|c_{n-1}|^2+...+|c_0|^2}\}$ and using the fact that in that open circle $g$ has $n$ roots ($0$ is zero of order $n$) and then get by the thorem that the polynomial $f$ has n roots in that circle. The problem is that I cant show that $|f(z)-g(z)|&lt;|g(z)|$ on the boundary of the circle. Can you please give me a hint how to show that?
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$\log x =Cx^4$ has only one root. Find C.  I don't know how to solve this problem. Do you take derivative on both sides?  I am thinking C equals 0. Am I correct on that?
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The question is:    Question.  Let $f,g:\mathbb R\to\mathbb R$ be two functions that satisfy$$f(x-y)=f(x)\cdot g(y)-f(y)\cdot g(x)$$ and $$g(x-y)=g(x)\cdot g(y)+f(x)\cdot f(y)$$ for all $x,y \in \mathbb{R} $.  If the right hand derivative at $x=0$ exists for $f(x)$, then what is $g'(0)$ ?  My try:  By some simple substitutions I figured out that $f(0)=0$ and $g(0)=1$. If in the second equation, we put $x=y$, it will give $g(0)=(g(x))^2+(f(x))^2$. If $g(0)=0$, sum of the two squares becomes $0$ which implies the squares themselves are zero, I neglected $g(x)=f(x)=0$ as a trivial solution and hence took $g(0)=1$. But how do I proceed after this?