What is the answer to 2^{2R-4}-5^{2-R}=0 ?

The answer to 2^{2R-4}-5^{2-R}=0 is R=2

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2^{2R-4}-5^{2-R}=0

2^{2 R - 4} - 5^{2 - R} = 0

R = 2

Solution steps

2^{2 R - 4} - 5^{2 - R} = 0

Add

5^{2 - R}

to both sides

2^{2 R - 4} - 5^{2 - R} + 5^{2 - R} = 0 + 5^{2 - R}

Simplify

2^{2 R - 4} = 5^{2 - R}

Apply exponent rules

(2 R - 4) ln (2) = (2 - R) ln (5)

Solve

(2 R - 4) ln (2) = (2 - R) ln (5) : R = 2

R = 2

4^{2 x} = 5^{2 x - 1}

6 4^{3 x} = 51 2^{2 x + 12}

8^{x + 2} = (\frac{1}{32})^{- 5 x}

\frac{1}{4 ^{x}} = \frac{1}{64}

e^{x + 2} = 3^{x}

What is the answer to 2^{2R-4}-5^{2-R}=0 ?
The answer to 2^{2R-4}-5^{2-R}=0 is R=2
Solve for R: 2^{2R-4}-5^{2-R}=0
The solution is R=2