How do you solve #4ln(5x)+5=2#?

2 Answers
Jul 6, 2017

#x = e^(-3/4)/5#

Explanation:

In order to solve this, we will begin with removing 5 from both sides:
#4ln(5x) + 5 - 5 = 2 - 5#.

Thus, we will get #4ln(5x) = -3#.

Then we will divide both sides by 4 to get #ln(5x) = -3/4#.

Then, to get x alone, we need to recognize the relationship of #e# and #ln#. If we rewrite both sides as #e^(ln(5x)) = e^(-3/4)# we can perform our next step.

Remember that #ln# and #e# are inverses of each other and that #ln(e) = 1#. Similarly, #e^ln(x) = x#. If you look at #e^ln(5x)#, you will recognize that we will get #5x# due to #ln# and #e# undoing each other.

Thus, continuing: #5x = e^(-3/4)#.

Divide both sides by 5 and we will get #x = e^(-3/4)/5#.

If you plug this into the equation for x, you will get #4ln(5(e^(-3/4)/5)) + 5 = 2#.

Jul 6, 2017

#x = e^(-3/4)/5#

Explanation:

Isolate the logarithm and then cancel it out by raising everything to a common base.

First, subtract 5 from both sides.

#4ln(5x)+5-5=2-5#

#4ln(5x)+cancel5-cancel5=2-5#

#4ln(5x) = -3#

Next, divide both sides by 4.

#4ln(5x) div 4 = -3 div 4#

#cancel4ln(5x) div cancel4 = -3 div 4#

#ln(5x) = -3/4#

Next, raise #e# to the power of both sides. This will cancel out the natural logarithm.

#e^ln(5x) = e^(-3/4)#

#cancele^(cancel"ln"(5x))=e^(-3/4)#

#5x = e^(-3/4)#

And finally divide by 5.

#5x div 5 = e^(-3/4)div5#

#cancel5x div cancel5 = e^(-3/4)/5#

#x = e^(-3/4)/5#

Final Answer