How do you solve #Log_2 (10X + 4) - Log_2 X = 3#?

1 Answer
Jan 5, 2016

No solution.

Explanation:

First, simplify using the rule that #log_a b-log_a c=log_a(b/c)#.

#log_2((10X+4)/X)=3#

To undo the logarithm, exponentiate both sides with base #2#.

#2^(log_2((10X+4)/X))=2^3#

#(10X+4)/X=8#

Solve for #X#.

#10X+4=8X#

#X=-2#

Warning! This is an invalid answer. If #X=-2#, then both of the logarithm functions in the original equation would have a negative argument. It's impossible to take the logarithm of a negative number.

If we graph this as a function, the graph should never cross the #x#-axis, indicating a lack of roots.

graph{ln(10x+4)/ln2-lnx/ln2-3 [-5.64, 22.84, -3.47, 10.77]}