((1/3)^{-1+log _{1/3} 4)^{1-x/3}=144

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вопрос

To solve the equation, we will simplify it step by step:

Start by rewriting the equation with the given values:
(1/3)^{-1+log _{1/3} 4)^{1-x/3}=144

Replace log{1/3}4 with its equivalent in exponential form:
(1/3)^{-1+(1/3)^{log {1/3} 4}}^{1-x/3}=144

Simplify the expression inside the brackets:
(1/3)^{-1+4}^{1-x/3}=144
(1/3)^{3}^{1-x/3}=144
(1/3)^{3(1-x/3)}=144
(1/3)^{3-3x/3}=144
(1/3)^{3-x}=144

Raise both sides to the power of 3:
(1/3)^{(3-x)*3}=144^3
1/3^{9-3x}=2985984

Simplify the equation further:
1/(3^9 * 3^(-3x)) = 2985984
1/(3^(9-3x)) = 2985984

Since 3^6 = 729, we know that 3^9 = 72927 = 19683:
1/19683 3^x = 2985984

Multiply both sides by 19683 to isolate 3^x:
3^x = 19683 * 2985984
3^x = 58823529472

Take the log base 3 of both sides to solve for x:
log_3(3^x) = log_3(58823529472)
x = log_3(58823529472)
x ≈ 23.911

Therefore, the value of x that satisfies the equation is approximately 23.911.