How do you solve #\ln ( 2x + 2) = 1.6#?

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Answer 1 · 2017-05-08T09:46:39

#x = e^1.6/2 - 1#

Firstly, identify that #color(red)ln# is called the natural logarithm.

This means it is a logarithm of a number N with a base #e#.

#log_e N# which can be expressed as #ln_e N#

Like #log_10 N# can be written as #lg N#, #ln_e N# can be written as #ln N# for short.

Therefore, #ln(2x+2) = 1.6#

Now, taking the exponential #e# for both sides, we get:

#color(red)e^ln(2x+2) = color(red)e^1.6#

Note: #e^color(red)(lna)= color(red)a#

#:.e^color(red)ln(2x+2) = color(red)(2x+2)#

Continuing,

#color(red)e^ln(2x+2) = color(red)e^1.6#

#2x+2 = e^1.6#

#2x = e^1.6 - 2# Dividing both sides by 2,

#x = e^1.6/2 - 1#