What is the answer to log_{2}(25^{x+3}-1)=2+log_{2}(5^{x+3}+1) ?

The answer to log_{2}(25^{x+3}-1)=2+log_{2}(5^{x+3}+1) is x=-2

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log_{2}(25^{x+3}-1)=2+log_{2}(5^{x+3}+1)

lo g_{2} (2 5^{x + 3} - 1) = 2 + lo g_{2} (5^{x + 3} + 1)

x = - 2

Solution steps

lo g_{2} (2 5^{x + 3} - 1) = 2 + lo g_{2} (5^{x + 3} + 1)

Apply exponent rules

lo g_{2} ((5^{x})^{2} \cdot 5^{6} - 1) = 2 + lo g_{2} (5^{x} \cdot 5^{3} + 1)

Rewrite the equation with

5^{x} = u

lo g_{2} ((u)^{2} \cdot 5^{6} - 1) = 2 + lo g_{2} (u \cdot 5^{3} + 1)

Solve

lo g_{2} (u^{2} \cdot 5^{6} - 1) = 2 + lo g_{2} (u \cdot 5^{3} + 1) : u = \frac{1}{25}

u = \frac{1}{25}

Substitute back

u = 5^{x},

solve for

x

Solve

5^{x} = \frac{1}{25} : x = - 2

x = - 2

\frac{2}{5} = (\frac{25}{4})^{x}

m, 144 = \frac{2 ^{m}}{3 ^{n}}

l, y^{m} - 6 y^{n} + 11 y^{l} - 6 y = 0

2 = (1.06)^{n}

2 7^{x} = 9

What is the answer to log_{2}(25^{x+3}-1)=2+log_{2}(5^{x+3}+1) ?
The answer to log_{2}(25^{x+3}-1)=2+log_{2}(5^{x+3}+1) is x=-2
Solve for x: log_{2}(25^{x+3}-1)=2+log_{2}(5^{x+3}+1)
The solution is x=-2