How do you solve #Log_(5)x + log_(3)x = 1#?

1 Answer
mason m
Feb 23, 2016

#x=e^((ln5(ln3))/(ln3+ln5))approx1.9211#

Explanation:

Use the change of base formula, which states that

#log_ab=log_cb/log_ca#

The common base I'll use here is #e#, since it is the base of the natural logarithm. It doesn't actually matter which base is chosen: any base will work.

The original expression can be rewritten as:

#lnx/ln5+lnx/ln3=1#

Find a common denominator of #ln5(ln3)#:

#(lnx(ln3))/(ln5(ln3))+(lnx(ln5))/(ln5(ln3))=1#

#(lnx(ln3)+lnx(ln5))/(ln5(ln3))=1#

Cross multiply.

#lnx(ln3)+lnx(ln5)=ln5(ln3)#

Factor a #lnx# from both terms on the left hand side.

#lnx(ln3+ln5)=ln5(ln3)#

Divide both sides by #ln3+ln5#.

#lnx=(ln5(ln3))/(ln3+ln5)#

To undo the natural logarithm, exponentiate both sides with base #e#.

#x=e^((ln5(ln3))/(ln3+ln5))approx1.9211#

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