Question

log base 5 of(x+1) + log base 5 of 2- log base 5 of (x-3) = log base 5 of (x-1)

how to solve

Answer 1

logbarcsec^(5)of(x+1)+logbarcsec^(5)of^(2)-logbarcsecof^(5)(x-3)=logbarcsec^(5)of(x-1)

Move all the terms containing a logarithm to the left-hand side of the equation.
logbarcsec^(5)of(x+1)+logbarcsec^(5)of^(2)-logbarcsecof^(5)(x-3)-logbarcsec^(5)of(x-1)=0

Multiply logb by each term inside the parentheses.
arcsec^(5)(x+1)logb+arcsec^(5)f^(2)ologb-logbarcsec(x-3)-logbarcsec^(5)(x-1)=0

Factor out the GCF of logb from each term in the polynomial.
logb(arcsec^(5)(x+1))+logb(arcsec^(5)f^(2)o)+logb(-arcsec(x-3))+logb(-arcsec^(5)(x-1))=0

Factor out the GCF of logb from arcsec(x+1)^(5)logb+logbarcsecf^(2)o^(5)-logbarcsec(x-3)-logbarcsec(x-1)^(5).
logb(arcsec^(5)(x+1)+arcsec^(5)f^(2)o-arcsec(x-3)-arcsec^(5)(x-1))=0

Multiply logb by each term inside the parentheses.
arcsec^(5)(x+1)logb+arcsec^(5)f^(2)ologb-arcsec(x-3)logb-arcsec^(5)(x-1)logb=0

Simplify -arcsec(x-3)logb by moving all terms inside the logarithm. The third law of logarithms states that the logarithm of a power of x is equal to the exponent of that power times the logarithm of x (e.g. log^b(x^(n))=nlog^b(x)).
arcsec^(5)(x+1)logb+arcsec^(5)f^(2)ologb+log(b^(-arcsec(x-3)))-arcsec^(5)(x-1)logb=0

Remove the negative exponent in the numerator by rewriting b^(-arcsec(x-3)) as (1)/(b^(arcsec(x-3))). A negative exponent follows the rule: a^(-n)=(1)/(a^(n)).
arcsec^(5)(x+1)logb+arcsec^(5)f^(2)ologb+log((1)/(b^(arcsec(x-3))))-arcsec^(5)(x-1)logb=0

This equation cannot be solved using the quadratic formula.
Can't Use Quadratic Formula