Solve 2 log x = log 64.

x = 8
x = ±8
x = 32
x = 128

Question

Solve 2 log x = log 64.

 x = 8
 x = ±8
 x = 32
 x = 128

anonymous · Answer

i think its b but im not sure

b is plus or minus 8

anonymous · Answer

2 $$2*32=64$$

mathmale · Answer

Please, would you show your work and mention which of the rules of logarithms you're using?  

Note that if you arrive at a particular answer, you can and should check it yourself by substituting the x value into the given equation.  Note also that -8 could not be a solution, because the domain of y = ln x is "all real numbers GREATER than zero."

mathmale · Answer

Rules of logs:

1.  log of a product:  ln x*y
2.  log of a quotient:  ln (x/y)
3.  log of a power:  ln (x^y)

anonymous · Answer

math male was my answer correct

anonymous · Answer

$$ \log _{a}(x^p) = p \log _{a} x $$
$$ \log (x^2) = 2 \log  x $$

mathmale · Answer

Purdy:  Please see my comments and suggestions, above, especially that involving x=-8.

mathmale · Answer

sangya21's suggestion is a very good one.  Can you apply it?

anonymous · Answer

Yes, thank you i got the answer

anonymous · Answer

Simplify 5 log2 k − 8 log2 m + 10 log2 n.

 7 log2 (k − m + n)
 7 log2 kn over m
 log2 50 kn over 8 m
 log2 k to the fifth power n to the tenth power over m to the eighth power

anonymous · Answer

can u help me with this question also?

mathmale · Answer

Great.
If you have another math problem you want to discuss, kindly enter it as a NEW question.

anonymous · Answer

do i have to type the whole thing if it dosen't paste?

anonymous · Answer

@mathmale

anonymous · Answer

Simplify
5 log2 k-8 log2 m+10 log2 n