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If two equivalent logarithmic expressions have the same base, then the arguments must be equal.
x=5
We want to solve an equation involving more than one logarithm. To do so, we will use the Product Property of Logarithms. log_b mn = log_b m + log_b n First, we will isolate the logarithm that contains the variable. Then, we can use the above property to isolate the variable from the logarithm. Recall that if the base is not stated, it is 10.
log(m) + log(n)=log(mn)
Distribute 5x
log(10^m)=m
Calculate power
Equate arguments
LHS-100=RHS-100
Rewrite - 5x as - 25x+20x
Factor out 5x & 20
Factor out (x-5)
Use the Zero Product Property
(I): LHS+5=RHS+5
(II): LHS-20=RHS-20
(II): .LHS /5.=.RHS /5.
To check for extraneous solutions, we will substitute both 5 and -4 for x in the given equation one at a time.
Since substituting 5 for x in the given equation produces a true statement, x=5 is the solution to our equation. Let's now check x=-4.
x= -4
a(- b)=- a * b
Subtract term
The argument of a logarithmic function cannot be negative, so both log(-20) and log(-5) are undefined, and -4 is an extraneous solution.