Question

solve the equation. log16^(3x-1)=log4^(3x)+log4^(0.5)
​

Answer 1

Answer:

Log[4](16)*log[16]4=1 where [] denotes base, is an example of a fundamental rule in logarithms: log[a](b)*log[b](a)=1, so, since log[4](16)=2log[4]4=2, log[16]4=1/2; or, alternatively, log[16]4=log[16]16^1/2=1/2log[16]16=1/2.

Also, log[a](x)=log[b](x)/log[b](a). Log[16](3x-1)=log[4](3x)+log[4](0.5)⇒2log[4](3x-1)=log[4](3x*0.5). We can equate the logs: (3x-1)^2=3x/2⇒9x^2-6x+1=3x/2⇒18x^2-12x-3x+2=0⇒

18x^2-15x+2=0⇒(3x-2)(6x-1)=0

From this x=2/3 or 1/6. Substitute these values in the original equation: 0=1/2-1/2. The value 1/6 cannot be used because it would require the log of a negative number so the only solution is x=2/3.

Step-by-step explanation: