How do you simplify $ln ((5 e^{x}) - (10 e^{2} x))$ $ln((5e^x)-(10e^2x))$ ?

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How do you simplify $ln ((5 e^{x}) - (10 e^{2} x))$ $ln((5e^x)-(10e^2x))$ ?

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Answer 1 · 2016-01-02T15:07:59

If you meant $ln ((5 e^{x}) - (10 e^{2 x}))$ $ln((5e^x)-(10e^(2x)))$

Then you can factor the $e^{x}$ $e^x$ and use $ln (a \cdot b) = ln a + ln b$ $ln(a*b)=lna+lnb$

$x + ln 5 + ln (1 - 2 e^{x})$ $x+ln5+ln(1-2e^x)$

Explanation:

It can't actually. You can't simplify polynomials with exponential functions. The fact that it is substraction (and not multiplication or division) leaves no room for simplifications.

However, if you meant $ln ((5 e^{x}) - (10 e^{2 x}))$ $ln((5e^x)-(10e^(2x)))$

$ln (5 e^{x} - 10 e^{x} \cdot e^{x})$ $ln(5e^x-10e^x*e^x)$

Factor the $5 e^{x}$ $5e^x$ :

$ln (5 \cdot e^{x} \cdot (1 - 2 e^{x}))$ $ln(5*e^x*(1-2e^x))$

Use of the property $ln (a \cdot b \cdot c) = ln a + ln b + ln c$ $ln(a*b*c)=lna+lnb+lnc$ gives:

$ln 5 + {ln e}^{x} + ln (1 - 2 e^{x})$ $ln5+lne^x+ln(1-2e^x)$

Since $ln = {log}_{e}$ $ln=log_e$

$ln 5 + x + ln (1 - 2 e^{x})$ $ln5+x+ln(1-2e^x)$

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How do you simplify $ln ((5 e^{x}) - (10 e^{2} x))$ $ln((5e^x)-(10e^2x))$ ?

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How do you simplify $ln ((5 e^{x}) - (10 e^{2} x))$ $ln((5e^x)-(10e^2x))$ ?