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

Question

Ariana Alvarez · Accepted Answer

No solution.

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]}

Serenity Jones · Answer

undefined To solve the equation Log_2(10X + 4) - Log_2(X) = 3, we can use the properties of logarithms. First, combine the logarithms using the quotient rule, which states that Log_a(b) - Log_a(c) = Log_a(b/c).

So, Log_2((10X + 4)/X) = 3.

Next, rewrite the equation in exponential form. For any base 'a', if Log_a(b) = c, then a^c = b.

So, 2^3 = (10X + 4)/X.

Now, solve for X:

2^3 = (10X + 4)/X 
=> 8 = (10X + 4)/X 
=> 8X = 10X + 4 
=> 8X - 10X = 4 
=> -2X = 4 
=> X = -2.

Thus, the solution to the equation Log_2(10X + 4) - Log_2(X) = 3 is X = -2.