What is the answer to log_{8}(x-1)-log_{8}(x-2)=1 ?

The answer to log_{8}(x-1)-log_{8}(x-2)=1 is x= 15/7

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log_{8}(x-1)-log_{8}(x-2)=1

lo g_{8} (x - 1) - lo g_{8} (x - 2) = 1

x = \frac{15}{7}

Solution steps

lo g_{8} (x - 1) - lo g_{8} (x - 2) = 1

Add

lo g_{8} (x - 2)

to both sides

lo g_{8} (x - 1) - lo g_{8} (x - 2) + lo g_{8} (x - 2) = 1 + lo g_{8} (x - 2)

Simplify

lo g_{8} (x - 1) = 1 + lo g_{8} (x - 2)

Apply log rules

x - 1 = 8 (x - 2)

Solve

x - 1 = 8 (x - 2) : x = \frac{15}{7}

x = \frac{15}{7}

Verify Solutions:

x = \frac{15}{7}

True

The solution is

x = \frac{15}{7}

lo g_{2} (x + 7) + lo g_{2} (x - 5) = 6

lo g_{9} (64 - x^{2}) = 1

- 3 ln (x + 2) = 0

lo g_{e} (6 - 8 x) = - 2

- 3 ln (x + 2) = 6

What is the answer to log_{8}(x-1)-log_{8}(x-2)=1 ?
The answer to log_{8}(x-1)-log_{8}(x-2)=1 is x= 15/7
Solve for x: log_{8}(x-1)-log_{8}(x-2)=1
The solution is x= 15/7