Question 384403
{{{3^(x+3)+3^x=84}}}
One way to solve this involves factoring in a way you may not have seen/done before. The first term on the left is the product of x+3 3's. The second term is the product of x 3's. So both terms have x 3's in common. We can factor out those x 3's from each term. After we factor out the x 3's, the first term will still have 3 3's. The second term will just be a 1 since we factored out all of its 3's:
{{{(3^x)(3^3+1)=84}}}
which simplifies to:
{{{(3^x)(28)=84}}}
Dividing both sides by 28 we get:
{{{3^x = 3}}}
So x = 1<br>
{{{4^(x+2)-4^x=15}}}
This problem can be solved in a similar way:
{{{(4^x)(4^2-1)=15}}}
{{{(4^x)(15)=15}}}
{{{4^x=1}}}
And so, if you remember how zero exponents work, x = 0.