# log 10 x - 1 = log 10

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log 10 x - 1 = log 10

Let us group similar:

==> log 10x - log 10 = 1

Now we know that log a - log b = log a/b

==> log 10x/10 = 1

==> log x = 1

==> x = 10^1

==> x = 10

We'll add 1 both sides:

log 10 x - 1 + 1 = log 10 + 1, but log 10 = 1

log 10x = log 10 + log 10

We'll use the product property, for the sum from the right side:

log 10x = log 10*10

log 10x = log 100

Because the bases of the logarithms are matching, we'll use the one to one property:

10x = 100

We'll divide by 10 to isolate x:

x = 10 > 0

Since x is positive, the equation has the solution x=10.

To solve log10x-1 = log10.

Solution:

log10x - 1= log10

log10x = 1+log10

log10x =log10+log 10, as log10 =1.

log10x = log10*10, as loga +logb = logab

lo10x = log 100. Take antilog.

x = 100/10 = 10.

10x = 100

The equation log 10 x - 1 = log 10 has to be solved for x.

Use the property of logarithm log a*b = log a + log b

log 10 + log x - 1 = log 10

Eliminate the log 10 from both the sides

log x - 1 = 0

Add 1 to both the sides

log x - 1 + 1 = 1

log x = 1

If log_b x = y, x = b^y

This gives: x = 10^1 = 10

The solution of the equation is x = 10