Solve for x: Log (base10)7 = log(base10)x - log(base10)2?

Answer 1

Let log be log to base 10 in the following : -
log 7 = log x - log 2
log x = log 7 + log 2
log x = log 14
x = 14

Answer 2

Add log 2 to both sides

log7+log2=logx
added logs is the same as multiplied numbers so:

log(7*2)=log x


14=x




x=14

Answer 3

Log 7 = log x - log2

Following properties of log... log(x/2) = log (x) - log (2)

Following yet another property of logs..

Log_10 7 = log_10 (x/2)

Remember, for log_10 7, 10 ^n = 7.. but 'n' in this cause is the log_10 (x/2)..

Also remember...

n ^(log_n x) --> is x..

For example..

log_3 9 = 2
3 ^2 = 9

3 ^ log_3(9) = 9.. this is because

log_3 (9) = 2

so 3 ^2 = 9


back to the problem...


10 ^ (log_10 (x/2)) = 7

(x/2) = 7

x = 14

Answer 4

log7 = logx - log2

using the second law of logs
(log a - b) = log(a/b)

log7 = log(x/2)

since the log of one thing is equal to the log of another thing then the two things must be equal

so 7 = x/2
then x = 14

Answer 5

log_10 (x) is the same as log(x). You don't have to include the 10

log(7) = log(x) - log(2)

add log(2) for both sides
log(7) + log(2) = log(x)

quotient rule: log(a) + log(b) = log(ab)
log(14) = log(x)

if log(a) = log(b) then a = b
x = 14

Answer 6

Since the basic question is already answered; I will just point out that :

logx implies a base 10 logarithm,

lnx implies base e and is called a natural logarithm

lgx implies base 2

anybody know any others?

Answer 7

all logs to base 10

log(7) = log(x) - log(2) = log(x/2)
exponentiate both sides
7 = x/2
x = 14

Answer 8

log7+log2=logx
log7*2=log x
14=x