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

Question

Sophie Adams · Accepted Answer

#x = -1/4#

Use the following logarithmic law first:

#ln (a/b) = ln(a) - ln(b)#

In your case, this leads to:

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

#<=> ln(e^(4x+3)) - ln(e) = 1#

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

Thus, #ln# and #e# eliminate each other in your equation, and you will get:

#<=> (4x + 3) - 1 = 1#

The solution of this equation is

#x = -1/4#

As #e^x# 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.

HIX Tutor · Answer

undefined Great question! Let’s break it down step-by-step.

1. **Look at the expression:** You have ln((e^(4x+3))/e) = 1.

2. **Simplify the inside of the natural log:** Recall that ln(a/b) = ln(a) - ln(b). So:
   $$
   ln((e^{(4x+3)})/e) = ln(e^{(4x+3)}) - ln(e)
   $$
   Since ln(e) = 1, we can rewrite it as:
   $$
   ln(e^{(4x+3)}) - 1
   $$

3. **Use the property of logarithms:** The property ln(e^y) = y lets us simplify further:
   $$
   4x + 3 - 1 = 1
   $$

4. **Combine like terms:** This gives us:
   $$
   4x + 2 = 1
   $$

5. **Isolate the variable:** Subtract 2 from both sides:
   $$
   4x = 1 - 2 \implies 4x = -1
   $$

6. **Solve for x:** Divide both sides by 4:
   $$
   x = -\frac{1}{4}
   $$

So, what do you think the value of $ x $ is?