How do you solve ln((e^(4x+3))/e)=1ln(e4x+3e)=1?

1 Answer
Lotusbluete
Dec 15, 2015

x = -1/4x=−14

Explanation:

Use the following logarithmic law first:

ln (a/b) = ln(a) - ln(b)ln(ab)=ln(a)−ln(b)

In your case, this leads to:

ln(e^(4x+3)/e) = 1ln(e4x+3e)=1

<=> ln(e^(4x+3)) - ln(e) = 1⇔ln(e4x+3)−ln(e)=1

As next, you need to use the property that ln xlnx and e^xex are inverse functions which means that ln(e^x) = xln(ex)=x and e^(ln x) = xelnx=x always hold.

Thus, lnln and ee eliminate each other in your equation, and you will get:

<=> (4x + 3) - 1 = 1⇔(4x+3)−1=1

The solution of this equation is

x = -1/4x=−14

As e^xex is always positive for any value of x in RR, and thus the logarithmic expression is defined for any x in RR, this is your solution.

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