1. Solving exponential equations

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?

2. $3(2^{x+1}) + 2(5^{x-2}) = 5^x + 2^{x-2}$ rearranging we get

$3(2^{x+1}) + 2^{x-2}= 5^x +2(5^{x-2})$ Now the laws of exponents tell us that $a^n + a^m = a^{m+n}$ so

$3(2^3)(2^{x-2}) + 2^{x-2} = (5^2)(5^{x-2}) + 2(5^{x-2})$ (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 $2^{x-2}$ and $5^{x-2}$ then take the log of both sides. Use your log laws to solve for x.

3. Originally Posted by Gusbob
Now the laws of exponents tell us that $a^n + a^m = a^{m+n}$
Just in case there is still some confusion from this, Gusbob meant to say $a^m \times a^n = a^{m+n}$

4. Originally Posted by Gusbob
$3(2^{x+1}) + 2(5^{x-2}) = 5^x + 2^{x-2}$ rearranging we get

$3(2^{x+1}) + 2^{x-2}= 5^x +2(5^{x-2})$
No. We get $3(2^{x+1}) - 2^{x-2}= 5^x - 2(5^{x-2})$
Now the laws of exponents tell us that $a^n + a^m = a^{m+n}$ so
$(a^n)(a^m)= a^{m+n}$ which is used correctly below

$3(2^3)(2^{x-2}) + 2^{x-2} = (5^2)(5^{x-2}) + 2(5^{x-2})$ (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 $2^{x-2}$ and $5^{x-2}$ then take the log of both sides. Use your log laws to solve for x.