3(2^(x+1)) + 2(5^(x-2)) = 5^x + 2^(x-2)
I know I'm supposed to move the base 2s and base 5s to one side and factor out 2^x and 5^x, but I have no idea how to do it. Any help?
3(2^(x+1)) + 2(5^(x-2)) = 5^x + 2^(x-2)
I know I'm supposed to move the base 2s and base 5s to one side and factor out 2^x and 5^x, but I have no idea how to do it. Any help?
$\displaystyle 3(2^{x+1}) + 2(5^{x-2}) = 5^x + 2^{x-2} $ rearranging we get
$\displaystyle 3(2^{x+1}) + 2^{x-2}= 5^x +2(5^{x-2}) $ Now the laws of exponents tell us that $\displaystyle a^n + a^m = a^{m+n} $ so
$\displaystyle 3(2^3)(2^{x-2}) + 2^{x-2} = (5^2)(5^{x-2}) + 2(5^{x-2}) $ (Note: It will almost always be easier to express in terms of the lowest power, in this case (x-2) for both exponents 2,5).
Now factorise $\displaystyle 2^{x-2} $ and $\displaystyle 5^{x-2} $ then take the log of both sides. Use your log laws to solve for x.
No. We get $\displaystyle 3(2^{x+1}) - 2^{x-2}= 5^x - 2(5^{x-2}) $
$\displaystyle (a^n)(a^m)= a^{m+n}$ which is used correctly belowNow the laws of exponents tell us that $\displaystyle a^n + a^m = a^{m+n} $ so
$\displaystyle 3(2^3)(2^{x-2}) + 2^{x-2} = (5^2)(5^{x-2}) + 2(5^{x-2}) $ (Note: It will almost always be easier to express in terms of the lowest power, in this case (x-2) for both exponents 2,5).
Now factorise $\displaystyle 2^{x-2} $ and $\displaystyle 5^{x-2} $ then take the log of both sides. Use your log laws to solve for x.