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Solve the exponential equation. Exact answers only 3e^(x-1)=2+e^(x-1)
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subtracting \(e^{x-1}\) from either side gives us what?
wait 3+2
or 3=2
@oldrin.bataku
:/ netta pretend the \(e^{x-1}\) is another variable like \(z\):$$3z=2+z$$how would you solve for \(z\) above?
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sub both sides by z
@oldrin.bataku
and you get \(2z=2\), yes? so \(z=1\))? well here we have \(z=e^{x-1}\) hence:$$e^{x-2}=1$$can you tell me what exponent makes a power yield \(1\)?
is it 1
@netta nah, \(0\):$$2^0=1\\3^0=1\\\dots$$ so we know the exponent of \(e^{x-1}\) (I made a typo above) needs to be \(0\):$$x-1=0\\x=1$$
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