$10e^{2x}-5=23e^x\qquad\text{subtract}\ 23e^x\ \text{from both sides}\\\\10e^{2x}-23e^x-5=0\\\\10(e^x)^2-23(e^x)-5=0\\\\\text{substitution:}\ e^x=t > 0\\\\10t^2-23t-5=0\\\\10t^2-25t+2t-5=0\\\\5t(2t-5)+1(2t-5)=0\\\\(2t-5)(5t+1)=0\iff2t-5=0\ \vee\ 5t+1=0\\\\2t=5\ \vee\ 5t=-1\\\\t=\dfrac{5}{2} > 0\ \vee\ t=-\dfrac{1}{5} < 0\\\\\text{therefore}\ e^x=\dfrac{5}{2}\to\ln e^x=\ln\left(\dfrac{5}{2}\right)\\\\\boxed{x=\ln\left(\dfrac{5}{2}\right)}$

5 0

3 years ago

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If log.. (125) = 3, then x =

Naddik [55]

5x5x5=125 so it could be x=5

6 0

3 years ago

How can I solve 4x-2>3x+5

cluponka [151]

You would move over the -2 making the equation look like this 4x>3x+7. Then move the 3x to the other side of the equation making it look like this 1x>7. Thus the answer is x>7.

7 0

3 years ago

Which best describes the life cycles of moss and angiosperm?

Aleksandr-060686 [28]

It would have to be iv

4 0

3 years ago

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