<p>Can anyone explain how to solve the following problems?</p>
<p>(x-1)^(log x-1 ) = 100(x-1)</p>
<p>log(base 2)X + log(base 4)X + log(base 8)X</p>
<p>4^X - 2^(X-1) = 3</p>
<p>(x-1)^(log x-1 ) = 100(x-1)</p>
<p>log of both sides</p>
<p>log(x-1)log(x-1)=log(100(x-1))</p>
<p>log(x-1)log(x-1)=log100+log(x-1)</p>
<p>call log(x-1)=A (which is a handy trick)</p>
<p>A^2=log100+A
A^2=2+A (assuming these are base 10 logs)
you should be able to solve for A from here and then figure out x.</p>
<p>log(base 2)X + log(base 4)X + log(base 8)X</p>
<p>change of base formula:</p>
<p>ln(x)/ln(2) + ln(x)/ln(4)+ln(x)/ln(8)</p>
<p>ln(x)/ln(2)+ln(x)/2ln(2)+ln(x)/3ln(2)</p>
<p>so, if the question asked you to simplify…</p>
<p>1/ln(2) * (lnx + 0.5lnx+0.333ln(x))</p>
<p>=log base 2 of x + log base 2 of sqrt(x) + log base 2 of cube root(x)</p>
<p>= log base 2 (x<em>x^0.5</em>x^0.33)=log base 2 (x^11/6)?</p>
<p>4^X - 2^(X-1) = 3 </p>
<p>4^x= 2^(2x) or (2^x)^2</p>
<p>and 2^(x-1) = 2^x*2^-1</p>
<p>(2^x)^2 - 0.5 * 2^x = 3</p>
<p>Again, 2^x = A</p>
<p>A^2-0.5A-3=0</p>