How to determine value of X when adding log with different bases?
log(base 16) x + log (base 4) x + log (base 2) x = 7
How do you solve it?
Well the answer is 16, but I want to know the solution.
Thank you.
ln(x)/ln(16)+ln(x)/ln(4)+ln(x)/ln(2)=7
ln(x)(1/ln(16)+1/ln(4)+1/ln(2))=7
ln(x)=7/(1/ln(16)+1/ln(4)+1/ln(2))
x=e^(7/(1/ln(16)+1/ln(4)+1/ln(2)))
Here, it helps to know that ln(16)=4ln(2) and ln(4)=2ln(2).
Remember, ln(a^b)=b*ln(a).
x=e^(7/(1/(4ln2)+1/(2ln2)+1/ln2))
x=e^(7/(7/(4ln2)))
x=e^(28ln2/7)
x=e^(4ln2)
x=e^(ln(2^4))
x=e^ln16
x=16
ln(x)/ln(16)+ln(x)/ln(4)+ln(x)/ln(2)=7
ln(x)(1/ln(16)+1/ln(4)+1/ln(2))=7
ln(x)=7/(1/ln(16)+1/ln(4)+1/ln(2))
x=e^(7/(1/ln(16)+1/ln(4)+1/ln(2)))
Here, it helps to know that ln(16)=4ln(2) and ln(4)=2ln(2).
Remember, ln(a^b)=b*ln(a).
x=e^(7/(1/(4ln2)+1/(2ln2)+1/ln2))
x=e^(7/(7/(4ln2)))
x=e^(28ln2/7)
x=e^(4ln2)
x=e^(ln(2^4))
x=e^ln16
x=16
References :
BAM.