<p>Can anyone help me out with this math problem? I keep getting stuck.
(3^x)+(3^-x)=10/3 thanx
Im more interested in the process than in the answer, so if you could somehow explain how you did it that would be awesome.</p>
<p>x=1</p>
<p>I kinda know how to solve it...but I really just thought about it for a second and it was obvious.</p>
<p>(3^x)+(3^-x)=10(3^-1)</p>
<h2>(3^x)+(3^-x)= 10</h2>
<pre><code> (3^-1)
</code></pre>
<p>3(3^x+3^-x)=10</p>
<p>(3^(x+1))+(3^(1-x))=10</p>
<p>And then I just realized that 3^2 + 3^0=10</p>
<p>so
x+1=2 (x=1)
1-x=0 (x=1)</p>
<p>And it works...</p>
<p>Is this SAT math? It seems somewhat difficult.</p>
<p>If you set y = 3^x, then the original equation becomes y + 1/y = 10/3
which gives you a quadratic equation to solve: y^2 -(10/3) y + 1 = 0
or 3y^2 - 10y + 3 = 0</p>
<p>Solve: y = {10 +/- sqrt(100 - 4(3)(3)) } / 6
= {10 +/- sqrt(64)} / 6
= 18/6 or 2/6
= 3 or 1/3</p>
<p>So 3^x = 3 => x=1
or 3^x = 3^-1 => x= -1</p>
<p>There are two solutions, -1 and 1 but I am stumped as to how to get it. I can only get one, and then I have been trying to multiply by 3^x and then use ln, but it doesnt work.</p>
<p>lil_killer, its not SAT math, its actually a problem I had for homework ( I'm in Math HL, but I'm just a junior, so that might explain why its not as difficult as other Mth Hl problems)</p>
<p>thank you all for your help</p>
<p>Were this question offered in a math competition (where analytical algebraic manipulations are not requrired), it could be approached this way:
y=3^x
10/3 = 3 1/3 = 3 + 1/3
y + 1/y = 3 + 1/3
y=3 or y=1/3
x=1 or x=-1.</p>
<p>Can someone either point out what is wrong w/ my math or does x just not exist? i'd just try graphing it on a calculator if you're allowed.</p>
<p>(3^x)+(3^-x)=10/3</p>
<p>which is the same as</p>
<p>(3^x) + (1/3^x) = 10/3</p>
<p>take the log of both sides to move the variable to the front</p>
<p>log3^x +log(1/3^x) = log(10/3)</p>
<p>that can be rewritten as</p>
<p>xlog3 + log1 - log (3^x) = log(10/3)</p>
<p>and simplified further to</p>
<p>xlog3 - xlog3 = log(10/3) - log(1) = log (10/3)</p>
<p>haha hmmm and then you get...</p>
<p>0 = log(10/3)</p>
<p>that doesn't work but why not....?</p>
<p>log ( a+b) = log a + log b</p>
<p>thank ha now i feel dumb. now i really wanna figure out that problem</p>
<p>Buy TI-89, F2--> solve, plug in function, solve for x?</p>
<p>X=2.45435567567546</p>
<p>The point of the problem seems to be more recognizing that 10/3 = 3/1 + 1/3. I sincerely doubt they'd ask you to solve this on the SAT or anywhere if it were any more complicated than that.</p>