What is the answer to 7^{(3-x)}=5^{(x+1)} ?

The answer to 7^{(3-x)}=5^{(x+1)} is x=(3ln(7)-ln(5))/(ln(35))

Solve for x: 7^{(3-x)}=5^{(x+1)}

The solution is x=(3ln(7)-ln(5))/(ln(35))

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7^{(3 - x)} = 5^{(x + 1)}

x = \frac{3 ln ( 7 ) - ln ( 5 )}{ln ( 35 )}

Decimal Notation

x = 1.18927 \dots

Solution steps

7^{(3 - x)} = 5^{(x + 1)}

Apply exponent rules

(3 - x) ln (7) = (x + 1) ln (5)

Solve

(3 - x) ln (7) = (x + 1) ln (5) : x = \frac{3 ln ( 7 ) - ln ( 5 )}{ln ( 35 )}

x = \frac{3 ln ( 7 ) - ln ( 5 )}{ln ( 35 )}

2 (3^{7 - 2 x}) + 48 = 102

2^{3 x - 13} = \frac{1}{2 ^{4}}

1.0 1^{n} = 2

ln (2^{3 x + 1}) = 5

e = 1 0^{x}

What is the answer to 7^{(3-x)}=5^{(x+1)} ?
The answer to 7^{(3-x)}=5^{(x+1)} is x=(3ln(7)-ln(5))/(ln(35))
Solve for x: 7^{(3-x)}=5^{(x+1)}
The solution is x=(3ln(7)-ln(5))/(ln(35))