Question 1072415
A) 9^(2x+3) =81^(2x-1)
<pre>
Write 81 as 9<sup>2</sup>

{{{9^(2x+3)=(9^2)^(2x-1)}}}

Multiply the exponents on the right
to rmove the parentheses:

{{{9^(2x+3)=9^(2(2x-1))}}}

Since the exponents on each side have
the same positive base other than 1,
The exponents are equal, so

{{{2x+3=2(2x-1)}}}

You finish.

</pre> 
B)2log(2x)=log(7x-3)
<pre>
{{{2*log((2x))=log((7x-3))}}}

Use the rule of logarithms that allows us
to move the 2 coefficent on the left up
above the (2x) as an exponent:

{{{log((2x)^2)=log((7x-3))}}}

Now what the log is of on the left must be
equal to what the log is of on the right,
so

{{{(2x)^2=7x-3}}}

You finish that.  It's a quadratic. 
Check both answers
</pre>
C)7*3^(x+2)=5x
<pre>
There is no way to solve an equation if the variable
appears both as part of exponent and also part of
a non-exponent, except as an approximation using a
graphing calculator.  However we can show with a
calculator that it has no solution because the left
side is always greater than the right side.

</pre>
d)log3(x+1)+log3(x+6)=2
<pre>
I can't tell whether the 3's are bases of the logarithms
or coefficients of the expressions in parentheses.
</pre>
E) log10^e
<pre>
"log" with no base written means that the base is 10.  So
{{{log(10,(10^e))}}} asks "What exponent of the base is
required to give what the log is of?"  What the log is of
is 10<sup>e</sup>, and so the exponent of the base 10 that
is required to give what the log is of, namely 10<sup>e</sup> is
simply the exponent e.

Edwin</pre>