<PRE><B>Question 721419</B>
First of all,
9^x^2 is the same as 9^(x^2) and (9^x)^2 (according to the power of a power rule for exponents) is the same as 9^(2x). As you can see, they are not the same. Since you did not provide any parentheses I&#39;m assuming that the first one is correct. If I&#39;m wrong then you will need to re-post your question (using parentheses like I did to make it clear).&lt;br&gt;
{{{3^(5x)*9^(x^2) = 27}}}
While any equation where the variable is in exponents can be solved using logarithms, some equations can be solved in another way. This other way&lt;ul&gt;&lt;li&gt;is easier&lt;/li&gt;&lt;li&gt;is faster&lt;/li&gt;&lt;li&gt;results in exact expressions for the solution. Using logarithms often involves the use of decimal approximations which can lead to small errors in the answers.&lt;/li&gt;&lt;li&gt;does not require a check. Solving with logarithms requires a check.&lt;/ul&gt;So it is definitely worth seeing if this other way to solve the equation can be used before you start using logarithms. If it is possible to write the equation so that both sides are powers of the same number/expression then this other way of solving the equation will work. In our equation, we have three bases. (The 27 has an exponent, too, even though we don&#39;t see it!) So we have to ask ourselves: &quot;Are 3, 9 and 27 all powers of the same number?&quot; If we can answer &quot;yes&quot;, then we can use the fast solution.&lt;br&gt;
We should recognize right away that 3 and 9 are both powers of 3. Is 27 a power of 3, too? With a little effort we should be able to figure out that {{{27 = 3^3}}}. So the answer to the big question is &quot;yes&quot;.&lt;br&gt;
The &quot;fast&quot; solution is:&lt;ol&gt;&lt;li&gt;Rewrite the equation so that each side is a power of the same number/expression.&lt;/li&gt;&lt;li&gt;Set the two exponents equal to each other. (The only way two powers of the same thing can be equal is if the exponents themselves are equal, too.&lt;/li&gt;&lt;li&gt;Solve the equation from step 2.&lt;/li&gt;&lt;/ol&gt;Now let&#39;s use this on our equation:
1. Rewrite.
First we replace 9 and 27 with powers of 3:
{{{3^(5x)*(3^2)^(x^2) = (3^3)}}}
Using the power of a power rule on {{{(3^2)^(x^2)}}} we get:
{{{3^(5x)*3^(2x^2) = (3^3)}}}
To multiply the powers of 3 on the left side we use the rule for exponents that tells us to add the exponents in this situation:
{{{3^(5x+2x^2) = (3^3)}}}
We now have both sides of the equation as powers of 3.&lt;br&gt;
2. Write an equation that says the exponents are equal.
{{{5x+2x^2 = 3}}}&lt;br&gt;
3. Solve.
We have a quadratic equation so we want it in standard form. Subtracting 3 and rearranging the terms we get:
{{{2x^2+5x-3 =0}}}
Now we factor (or use the quadratic formula). This factors without too much difficulty:
{{{(2x-1)(x+3) = 0}}}
From the Zero Product Property:
2x-1 = 0 or x+3 = 0
Solving these we get:
x = 1/2 or x = -3
Both of these are solutions to your original equation. But the problem asks for just the larger one. So x = 1/2 is the solution to your problem.</PRE>