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# Find the solution of the exponential equation

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Find the solution of the exponential equation

70.9x = 6

9−x/200 = 2

e2x + 1 = 900

Guest Jul 9, 2014

#3
+19207
+5

e2x + 1 = 900   ?

$$\begin{array}{rcrc} e^{2x+1} &=& 900 & \quad | \quad \ln{}\\ \ln{(e^{2x+1} )}&=&\ln{( 900 )} & \\ (2x+1)*\ln{( e )}&=&\ln{( 900 )} & \quad |\quad \ln{(e)} = 1 \quad !\\ 2x+1 &=&\ln{( 900 )} & \\ 2x&=&\ln{( 900 )} - 1& \\ x&=&{ \ln{(900)} -1 \over 2 } &\\ x&=&2.90119738166&\end{array}$$

heureka  Jul 10, 2014
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#1
+19207
+5

70.9x = 6   ?

$$\begin{array}{rcrc} 7^{0.9x} &=& 6 & \quad | \quad \ln{}\\ \ln{( 7^{0.9x} )} &=& \ln{( 6 )} & \\ 0.9*x*\ln{( 7 )} &=& \ln{( 6 )} & \\ x&=&{ \ln{(6)} \over 0.9 *\ln{(7)} } &\\ x&=&1.02309135685& \end{array}$$

heureka  Jul 10, 2014
#2
+19207
+5

9−x/200 =  ?

$$\begin{array}{rcrc} 9^{ (-{x\over 200}) } &=& 2 & \quad | \quad \ln{}\\ \ln{( 9^{ (-{x\over 200}) } ) }&=&\ln{(2 )} & \\ (-{x\over 200}) } *\ln{(9 )}&=&\ln{(2 )} & | \quad *\quad -({200\over \ln{(9)}})\\ x&=&-200*{\ln{(2)}\over\ln{(9)}} &\\ x&=&-63.0929753571& \end{array}$$

heureka  Jul 10, 2014
#3
+19207
+5
$$\begin{array}{rcrc} e^{2x+1} &=& 900 & \quad | \quad \ln{}\\ \ln{(e^{2x+1} )}&=&\ln{( 900 )} & \\ (2x+1)*\ln{( e )}&=&\ln{( 900 )} & \quad |\quad \ln{(e)} = 1 \quad !\\ 2x+1 &=&\ln{( 900 )} & \\ 2x&=&\ln{( 900 )} - 1& \\ x&=&{ \ln{(900)} -1 \over 2 } &\\ x&=&2.90119738166&\end{array}$$