Question 916604
<pre>
Must all be rounded to four decimal places 
{{{9e^(19x) = 16}}}

{{{ln(9e^(19x)) = ln(16)}}}

{{{ln(9)+ln(e^(19x))=ln(16)}}}

{{{ln(9)+19x=ln(16)}}}

{{{19x=ln(16)-ln(9)}}}

{{{x=(ln(16)-ln(9))/19}}}

{{{x=0.0303}}}

You got that one correct

{{{5(1+10^(2x)) = 8  
{{{5 + 5*10^(2x)= 8}}}
{{{5*10^(2x)=3}}}
{{{log(5)+log(10^(2x))=log(3)}}}
{{{log(5)+2x=log(3)}}}
{{{2x=log(3)-log(5)}}}
{{{x = (log(3)-log(5))/2}}}
{{{x = -.1109}}}

Oh! oh!, you missed that one.  Use "log" not "ln"

{{{10^(1-x) = 4x}}}

That one cannot be solved using algebra.  You can only
do it with a graphing calculator.  That's because if
a variable is both part of an exponent and also part od
something that is not an exponent, then there is no algebraic
method for solving it.  The calculator can do, but it
essentially does by trial and error, trying answers that get
closer and closer to solving it and giving the closest one
it can find. 

With a graphing calculator, the solution is

x = 0.6115 

-----------------------------

{{{e^x - 12e^x - 1 = 0}}}

The first two terms on the left are like terms,
(since both have e<sup>x</sup> ans 1-12 = -11)

{{{1e^x - 12e^x - 1 = 0}}}
{{{-11e^x - 1 = 0}}}
{{{-11e^x = 1}}}
{{{e^x = -1/11}}}

There is no solution because "e" raised to any power is 
always positive and therefore can never equal a negative 
number.  Did you copy the problem wrong?

Edwin</pre>