Answer:
log7(x) = log4(x)
Change of base formula:
logA x = logB x / logB a
Changing to base E:
log7(x) = logE(x)/logE(7) = ln(x)/ ln(7)
log4(x) = lonE(x)/logE(4) = ln(x)/ln(4)
So they are equal when :
ln(x)/ln(7) = ln(x)/ln(4)
cross multplying:
ln(x)*ln(4) = ln(x)*ln(7)
ln(x)*ln(4) - ln(x)*ln(7) = 0 <--- everybody to left side
ln(x) * [ ln(4) - ln(7)] = 0 <--- factors out ln(x)
Let k = [ ln(4) - ln(7) ] is a fixed number constant.
Then,
ln(x) * k = 0
dividing both sides by k, cancels it out.
ln(x) = 0 whieans x=1.
So they are equal when x=1
Check:
For x=
log7(1) = log4(1)
is the same as asking the question:
7^x = 1 = 4^x are both equal to 1 where x is the
same exponent in both....
Of course, any number to power zer0 is 1. So x=1.
The graph in
