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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

Best Answer 

 #3
avatar+19630 
+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
 #1
avatar+19630 
+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
avatar+19630 
+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
avatar+19630 
+5
Best Answer

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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