How do you solve #e^(12-5x) -7 = 123#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Sudip Sinha Jul 30, 2015 # x = 1/5 (12 - ln(130)) # Explanation: # e^(12−5x)−7 = 123 # # => e^(12−5x) = 123 + 7 = 130 # (Add 7 to both sides) # => 12−5x = ln(130) # (Take natural logarithm on both sides) # => −5x = -12 + ln(130) # (Add -12 to both sides) # => 5x = 12 - ln(130) # (Multiply by -1) # => x = 1/5 (12 - ln(130)) # (Divide by 5 on both sides) If you need, you may use a calculator to get # log(130) = 4.8675 #. Put this value to get # x = 1.4265 #. Answer link Related questions What is a logarithmic model? How do I use a logarithmic model to solve applications? What is the advantage of a logarithmic model? How does the Richter scale measure magnitude? What is the range of the Richter scale? How do you solve #9^(x-4)=81#? How do you solve #logx+log(x+15)=2#? How do you solve the equation #2 log4(x + 7)-log4(16) = 2#? How do you solve #2 log x^4 = 16#? How do you solve #2+log_3(2x+5)-log_3x=4#? See all questions in Logarithmic Models Impact of this question 2745 views around the world You can reuse this answer Creative Commons License