**Q: Solve **

log_{16}x + log_{4}x + log_{2}x = 7

A: It is easier (though not always necessary) to have all of the logs in the same base to proceed. Since that is not the case, we need to use the change of base formula to put all of the logs into the same base.

Change of base formula says:

log_{a}b = log_{c}b / log_{c}a *where c can be any base of your choosing*

What base should we go to? Since 16, 4 and 2 can all be formed by powers of 2, let’s go to base 2:

log_{16}x + log_{4}x + log_{2}x = 7

log_{2}x / log_{2}16 + log_{2}x / log_{2}4 + log_{2}x / log_{2}2 = 7

Now, we can simplify:

log_{2}x / 4 + log_{2}x / 2 + log_{2}x / 1 = 7

1/4 log_{2}x + 1/2 log_{2}x + log_{2}x = 7

7/4 log_{2}x = 7

log_{2}x = 7 *(4/7)

log_{2}x = 4

(remember, this reads: *the power you put on 2 to get x is 4)*

x = 2^{4}

x = 16