Question

How do you solve `log_10 7=log_10 x-log_10 2`?

Answer 1

`x=14`

Explanation:

For two logarithms of the same base, `log_a M=log_a N rArr M=N`

Using the Quotient Law, `log_a x-log_a y=log_a (x/y)`

So,

`log_10 7= log_10 (x/2)`
`7=x/2`
`x=7*2`
`x=14`

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ABR May I suggest another approach:

Logs are indices in another form as such adding indices is another representation of multiplication. Like wise, subtraction is another representation of division. Consequently:

It is given that `log7 = log x - log 2`

( it does not matter a bout what the base is in this context)

`Antilog( log 7) = 7, Antilog( log x) =x, Antilog( log 2) =2`

Thus `log 7 = log x - log 2 " is equivalent to " 7 = x/2`

From this it is a simple matter of manipulation to find `x`