Writing $\int_{-\infty}^{a} b\cdot f(x)dx$ as $\int_{-\infty}^{c} f(x)dx$ for some $c$, for $f$ a gaussian

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For $\frac{1}{2}<b<1, a\in \mathbb{R}$ given, and $f(x)=\frac{1}{2\pi}e^{-x^2/2}$, I want to find $c$ such that $\int_{-\infty}^{a} b\cdot f(x)dx=\int_{-\infty}^{c} f(x)dx$. I might also be ok with the RHS being integral over a gaussian with different mean or variance.

My work so far:

  1. It should be possible, as the integrals are continuous functions (with variable the upper bound of the integration domain), and so the intermediate value theorem gives existance since $\int_{-\infty}^{-\infty} f(x)dx\leq \int_{-\infty}^{a} b\cdot f(x)dx \leq \int_{-\infty}^{\infty} f(x)dx$.
  2. Linear change of variables don't work, to the best of my understanding. At lease I tried $y=b^d x$ and saw that in order to get a gaussian in the RHS we must have $-(1-d)=d$ which can't be.