I cannot figure out this logarithm! 8^(2x+1)=16^(1-x)
I rewrote the coefficients to have the same base...2^3 and 2^4, what do i do next?? 2^3(2x+1)=2^4(1-x) do i factor both sides or whatt?? :(

Question

I cannot figure out this logarithm! 8^(2x+1)=16^(1-x)
I rewrote the coefficients to have the same base...2^3 and 2^4, what do i do next?? 2^3(2x+1)=2^4(1-x) do i factor both sides or whatt?? :(

anonymous · Answer

$$\Large 8^{2x + 1} = 16^{1 - x}$$
$$\Large 2^{6x + 3} = 2^{4 - 4x}$$
Divide both sides by 2^(4 -4x)
$$\Large \frac{2^{6x+3}}{2^{4 - 4x}} = 1$$
$$\Large 2^{6x + 3 - (4 - 4x)} = 2^{10x - 1} = 1$$
$$\Large 2^{10x} = 2^1$$
$$10x = 1$$
$$x = \frac{1}{10}$$

dumbcow · Answer

since they have same base .... take log (base 2) of both sides leaving

--> 3(2x+1) = 4(1-x)

anonymous · Answer

^ better way than mine

anonymous · Answer

i just had a serious math meltdown. thank you both very much for clearing that up!!!