= Solution
The exponents are real, as required for this family of real <continuous functions>. We prove the <exponential Chebyshev system> property by induction on the number of <functions>. A single nonzero multiple of an <exponential function> has no zero, so the base case holds.
Suppose the result holds for $n$ distinct real exponents, and consider a nontrivial combination $F(x)=\sum_{j=0}^n c_je^{\lambda_jx}$. If it had $n+1$ distinct zeros, so would
$$
g(x)=e^{-\lambda_0x}F(x)=c_0+\sum_{j=1}^n c_je^{(\lambda_j-\lambda_0)x},
$$
since the prefactor is everywhere positive. Applying <Rolle's theorem> between consecutive zeros gives at least $n$ distinct zeros of
$$
g'(x)=\sum_{j=1}^n c_j(\lambda_j-\lambda_0)e^{(\lambda_j-\lambda_0)x}.
$$
The $n$ exponents in this <derivative> are distinct. By induction, a nontrivial combination of them has at most $n-1$ zeros. Therefore every coefficient $c_j(\lambda_j-\lambda_0)$ must be zero. Distinctness gives $c_j=0$ for $j\ge1$. Then $g=c_0$ has a zero only if $c_0=0$, contradicting the original nontriviality. Consequently $F$ has at most $n$ zeros, and
$$
\boxed{(e^{\lambda_0x},\ldots,e^{\lambda_nx})\text{ is a Chebyshev system on }[a,b].}
$$
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