<p>I'm doing some Algebra II hw... and these problems are giving me migraines. Their impossible... my teacher gave the hint that one must use substitution. Well here are two of them...</p>
<p>2^x - 3^x = 0 It looks so simple but I can't figure it out...</p>
<p>2^(2/3)x+1 - 3 . 2^(1/3)x - 20 = 0 Confusing.... and the . is multiply dot.</p>
<ol>
<li><p>2^x=3^x
this is true iff (if and only if) x = 0 (and therefore 2^x=3^x=1)</p></li>
<li><p>Could you clarify the order of operations? Is the +1 part of the first exponent? Is the -20 part of the second exponent?</p></li>
</ol>
<p>1) 2^x-3^x = 0
2^x=3^x
x=0</p>
<p>Yea its 0, cuz 0 as an exponents yields 1 for nearly any number.</p>
<p>ok this second one could be wrong.</p>
<p>2)</p>
<p>2^(2/3)x+1 - 3 . 2^(1/3)x - 20 = 0
u=2^x/3</p>
<p>u^2 - 3u - 19 = 0</p>
<p>u = (3 (+/-) sqrt(85)) / 2</p>
<p>2^x/3 = (3 (+/-) sqrt(85)) / 2</p>
<p>x/3 = log(base 2) ((3 (+/-) sqrt(85)) / 2)</p>
<hr>
<p>x = 3log(base 2) ((3 (+/-) sqrt(85)) / 2)</p>
<p>Thanks! I get number one now! ^^ But the +1 part is part of the exponent for the second one not a seperate operation, but -20 is not part of the other exponent. The answer is supposed to be 6, but I have no idea how to get that.</p>
<p>Okay, I figured it out. The +1 is not multiplied by the 2/3. It's 2^((2x/3)+1) That makes it work (thanks for filling me in on the answer, that's how I was able to work it out).</p>
<p>So here's how to solve it: use the u-method above, but leaving the treating the +1 in the first exponent as *2^1, so that you get:</p>
<p>(2u^2)-(3u)-(20)=0</p>
<p>Factor that, and you'll get u = 4 or -2/5. Throw out -2/5 b/c exponentials are never negative. That leaves you with 2^(x/3)=4=2^2</p>
<p>If 2^(x/3)=2^2, x/3=2. That means that x must equal 6.</p>
<p>ah makes sense. I think mine may have been correct with the misinterpretation, but it just seemed too random an answer for a textbook problem.</p>
<p>I get most of it, but where did the +1 go? I didn't understand that part</p>