How do you solve #ln(4x+1)-ln3=5#?

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Answer 1 · 2016-11-11T18:13:39

Please see the explanation for steps leading to: #x = (3e^5 - 1)/4#

Use the property of logarithms #log_b(x) - log_b(y) = log_b(x/y)#:

#ln((4x + 1)/3) = 5#

Write both sides as exponents of e:

#e^{ln((4x + 1)/3)} = e^5#

Use the property #e^ln(a) = a#:

#(4x + 1)/3 = e^5#

Multiply both side by 3:

#4x + 1 = 3e^5#

Subtract 1 from both sides:

#4x = 3e^5 - 1#

Divide both sides by 4:

#x = (3e^5 - 1)/4#