3-5-2x-1-7-2-4x-1-19-x-

Question Number 181466 by SEKRET last updated on 25/Nov/22

3 \cdot 5^{2 x + 1} - 7 \cdot 2^{4 x + 1} = 19

x = ?

Answered by MJS_new last updated on 25/Nov/22

3 \times 5^{2 x + 1} - 7 \times 2^{4 x + 1} = 19

15 \times 25^{x} - 14 \times 16^{x} = 19

well \dots obviously

15 \times 5 - 14 \times 4 = 19

\Rightarrow

x = \frac{1}{2}

15 \times 25^{x} - 14 \times 16^{x} = 19

we can try to solve

15 p - 14 q = 19

with p > 0 \land q > 0 and see if we find a pair

(p, q) with \frac{\ln p}{\ln 25} = \frac{\ln q}{\ln 16}

p = 14 n + 5 \land q = 15 n + 4; n \in N

we get

p = 5 \land q = 4 ★

p = q = 19

p = 14 \land q = 34

\dots

but because 15 \times 25^{x} - 14 \times 16^{x} is strictly

increasing for x > 0 (x >\approx - . 489021) there

is only one solution

Commented by SEKRET last updated on 25/Nov/22

thanks sir!

Facebook Tweet Pin