+26388 Find all values of x for which (4x-5) to the power of 2x+5=1.
\(\begin{array}{|rclcrcll|} \hline && (4x-5)^{(2x+5)} &=& 1 && \qquad & | \qquad \ln() \\ && \ln( (4x-5)^{(2x+5)} ) &=& \ln(1) && \qquad & | \qquad \ln(1) = 0 \\ && \ln( (4x-5)^{(2x+5)} ) &=& 0 && \qquad & | \qquad \ln(a^b) = b\cdot \ln(a) \\ && (2x+5) \cdot \ln( 4x-5 ) &=& 0 \\ (2x+5) &=& 0 & or &\qquad \ln( 4x-5 ) &=& 0 \\ 2x &=& -5 & or &\qquad e^{\ln( 4x-5 )} &=& e^{0} \\ \mathbf{ x } & \mathbf{=} & \mathbf{-\frac52} & or &\qquad 4x-5 &=& 1 \\ && & &\qquad 4x &=& 6 \\ && & &\qquad x &=& \frac64 \\ && & &\qquad \mathbf{ x } & \mathbf{=} & \mathbf{\frac32} \\ \hline \end{array}\)
+37093 To raise something to a 'POWER' and get ' 1 ' the exponent must be '0'
so 2x+5 MUST = 0 2x+5 = 0 means x= -5/2
+26388 Find all values of x for which (4x-5) to the power of 2x+5=1.
\(\begin{array}{|rclcrcll|} \hline && (4x-5)^{(2x+5)} &=& 1 && \qquad & | \qquad \ln() \\ && \ln( (4x-5)^{(2x+5)} ) &=& \ln(1) && \qquad & | \qquad \ln(1) = 0 \\ && \ln( (4x-5)^{(2x+5)} ) &=& 0 && \qquad & | \qquad \ln(a^b) = b\cdot \ln(a) \\ && (2x+5) \cdot \ln( 4x-5 ) &=& 0 \\ (2x+5) &=& 0 & or &\qquad \ln( 4x-5 ) &=& 0 \\ 2x &=& -5 & or &\qquad e^{\ln( 4x-5 )} &=& e^{0} \\ \mathbf{ x } & \mathbf{=} & \mathbf{-\frac52} & or &\qquad 4x-5 &=& 1 \\ && & &\qquad 4x &=& 6 \\ && & &\qquad x &=& \frac64 \\ && & &\qquad \mathbf{ x } & \mathbf{=} & \mathbf{\frac32} \\ \hline \end{array}\)
+26388 Find all values of x for which (4x-5) to the power of 2x+5=1.
\(\begin{array}{|rclcrcll|} \hline && (4x-5)^{(2x+5)} &=& 1 && \qquad & | \qquad \ln() \\ && \ln( (4x-5)^{(2x+5)} ) &=& \ln(1) && \qquad & | \qquad \ln(1) = 0 \\ && \ln( (4x-5)^{(2x+5)} ) &=& 0 && \qquad & | \qquad \ln(a^b) = b\cdot \ln(a) \\ && (2x+5) \cdot \ln( 4x-5 ) &=& 0 \\ (2x+5) &=& 0 & or &\qquad \ln( 4x-5 ) &=& 0 \\ 2x &=& -5 & or &\qquad e^{\ln( 4x-5 )} &=& e^{0} \\ \mathbf{ x } & \mathbf{=} & \mathbf{-\frac52} & or &\qquad 4x-5 &=& 1 \\ && & &\qquad 4x &=& 6 \\ && & &\qquad x &=& \frac64 \\ && & &\qquad \mathbf{ x } & \mathbf{=} & \mathbf{\frac32} \\ \hline \end{array}\)
So the exponent must be zero => 2x+5 = 0 => x = -5/2
or the base must be 1 => 4x-5 = 1 => x = 3/2
+37093 Good point Heureka....guess we can all save a lot of computations by realizing either
1 The base must = 1 so 4x-5 =1 x= 6/4 = 3/2
or
2 The exponent must = 0 so 2x+5=0 x = - 5/2
Simple !
Solve for x over the real numbers:
(4 x-5)^(2 x+5) = 1
Take the natural logarithm of both sides:
log(4 x-5) (2 x+5) = 0
Split log(4 x-5) (2 x+5) into separate parts with additional assumptions.
Assume 4 x-5!=0 from log(4 x-5):
2 x+5 = 0 for 4 x-5!=0
or log(4 x-5) = 0
Subtract 5 from both sides:
2 x = -5 for 4 x-5!=0
or log(4 x-5) = 0
Divide both sides by 2:
x = -5/2 or log(4 x-5) = 0
Cancel logarithms by taking exp of both sides:
x = -5/2 or 4 x-5 = 1
Add 5 to both sides:
x = -5/2 or 4 x = 6
Divide both sides by 4:
Answer: |x = -5/2 or x = 3/2
