How do you solve #ln(5-2x^2)+ln9=ln43#?

3 Answers
Sonnhard · Jacobi J.
May 27, 2018

#x=pm1/3#

Explanation:

We have
#ln(5-2x^2)=ln(43)-ln(9)#

#ln(5-2x^2)=ln(43/9)#

so

#5-2x^2=43/9#

#5-43/9=2x^2#

#2/9=2x^2#

#1/9=x^2#

F. Javier B.
May 27, 2018

See explanation below

Explanation:

The goal is to get a expresion #logA=logB#. By inyectivity of logarithm we arrive to #A=B#

In our case: #ln(5-2x^2)+ln9=ln43#

#ln(5-2x^2)=ln43-ln9=ln43/9#

Then #5-2x^2=43/9# or equivalent

#5-43/9=2x^2#

#2/9=2x^2#

#1/9=x^2#

#x=+-1/3#

It is obvious that both solution are valid

marfre
May 27, 2018

#x = +- 1/3#

Explanation:

Given: #ln (5 - 2x^2) + ln 9 = ln 43#

Use the logarithm property #ln a + ln b = ln (a*b)#

#ln ((5 - 2x^2)9) = ln 43#

#ln (45 - 18x^2) = ln 43#

Exponentiate both sides and use the logarithm property #e^ln x = x#

#e^(ln (45 - 18x^2)) = e^(ln 43)#

#45 - 18x^2 = 43#

#45 - 43 = 18x^2#

#2 = 18 x^2#

#2/18 = 1/9 = x^2#

#x = +- 1/3#

CHECK to see if the answers work in the problem (must be ln of a positive number)#:

#ln (5-2(1/3)^2) = ln (5-2/9) > 0#

#ln (5-2(-1/3)^2) = ln (5-2/9) > 0#

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