How do you simplify #log (x+6)=1-log(x-5)#?

2 Answers
Gió
May 11, 2018

I tried this:

Explanation:

We can use, as an example, the base of the logs as being #10#. We get:

#log_(10)(x+6)=1-log_(10)(x-5)#

rearrange:

#log_(10)(x+6)+log_(10)(x-5)=1#

use a property of logs to write:

#log_(10)[(x+6)(x-5)]=1#

use the definition of log to write:

#(x+6)(x-5)=10^1#

that becomes:

#x^2-5x+6x-30-10=0#
#x^2+x-40=0#

Use the Quadratic Formula:

#x_(1,2)=(-1+-sqrt(1+160))/2=(-1+-sqrt(161))/2#
and:
#x_1=(-1+sqrt(161))/2=5.8442#
#x_2=(-1-sqrt(161))/2=-6.8442#

the negative solution cannot be accepted.

Mr. Hsia
May 11, 2018

#x = (-1 + sqrt(161))/2 ~~ 5.844#

Explanation:

#log(x+6) = 1 - log(x-5)#

Move #log(x-5)# to Left Side
#log(x+6) + log(x-5) = 1#

Condense the Left Side into 1 Log expression
#log((x+6)(x-5)) = 1#

Convert from Log Equation to Exponential Equation
#(x+6)(x-5) = 10^1#

Multiply Left Side
#x^2 - x - 30 = 10#

Move #10# to Left Side
#x^2 - x - 40 = 0#

Since we can't factor, use Quadratic Formula:
Quadratic Formula: #x = (-b +- sqrt(b^2 - 4ac))/(2a)#
#x = (-1 + sqrt(161))/2 ~~ 5.844#
And
#x = (-1 - sqrt(161))/2 ~~ -6.844#

Since the "inside" of log has to be positive,
#x = (-1 + sqrt(161))/2 ~~ 5.844# is the answer.

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