What is ln(e(5x))+e^0 ?

The solution to ln(e(5x))+e^0 is ln(x)+2+ln(5)

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simplify ln(e(5x))+e^0

simplify

ln (e (5 x)) + e^{0}

ln (x) + 2 + ln (5)

Solution steps

ln (e (5 x)) + e^{0}

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

ln (e (5 x)) = ln (e) + ln (5 x)

= ln (e) + ln (5 x) + e^{0}

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

ln (5 x) = ln (5) + ln (x)

= ln (e) + ln (5) + ln (x) + e^{0}

Apply rule

a^{0} = 1, a \neq = 0

e^{0} = 1

= ln (e) + ln (5) + ln (x) + 1

Apply log rule:

lo g_{a} (a) = 1

= 1 + ln (5) + ln (x) + 1

Group like terms

= ln (x) + ln (5) + 1 + 1

Add the numbers:

1 + 1 = 2

= ln (x) + 2 + ln (5)

lo g_{e} (x x^{2} + 1)

- 0.6 \cdot lo g_{2} (0.6) - 0.4 \cdot lo g_{2} (0.4)

- lo g_{10} (6.7 \cdot 1 0^{- 8})

ln (\frac{153.5}{100}) + (\frac{1}{2} \cdot 0. 4^{2} - 0.02) \cdot 4

\frac{1}{2} ln (2 x)