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[@SidneiMV]

(11/4)^(x + 5) = 1 x = -5

[@olejohaneikanger3715]

As long as you have two different values raised to the same exponent equalling each other, the only exponent to make that correct is 0. So x + 5 must be 0, making x = -5.

[@spdas5942]

Clearly x+5=0 => x=-5

[@petercrem-wl9ev]

More pedantic arithmetic.

[@gregevgeni1864]

4^(x+5) = 11^(x+5) => (4/11)^(x+5) =1 = (4/11)⁰ => x+5 =1 => x=-4 😂😂 Ι apologize ..x+5 = 0 => x=-5 ...

[@prollysine]

2x+10=(log11/log2)*x+(log11/log2)*5 , 10-(log11/log2)*5=(log11/log2)*x-2x , 5*(2-log11/log2)=x*((log11/log2)-2) , x=5*(2-log11/log2)/((log11/log2)-2) , x=5*(2-log11/log2)/(-(2-(log11/log2))) , x= -5 , test , 2^(2x+10)=1 , 11^(x+5)=1 , --> 2^0=11^0 , same , OK ,