(4^2x)+(4^x)-6=0
Solve the equation. Express the solution set in terms of natural logarithms. then use a calculator to obtain a decimal approximation to two decimal places.

Question

(4^2x)+(4^x)-6=0
Solve the equation. Express the solution set in terms of natural logarithms. then use a calculator to obtain a decimal approximation to two decimal places.

tkhunny · Answer

How would you solve this one?

$w^{2} + w - 6 = 0$

anonymous · Answer

factor?

tkhunny · Answer

Let's see.

anonymous · Answer

-3 and 2

anonymous · Answer

but it has to be in terms of Ln

tkhunny · Answer

It was more the process I was hoping to see.  Your work is far more valuable than the result.

$w^{2}+w−6=0$

$(w+3)(w-2) = 0$

$w = -3\;or\;w = 2$

Perfect.  Now, do EXACTLY the same thing with your original equation.

$(4^{2x})+(4^x)-6=0$

$(4^{x})^{2}+(4^x)-6=0$

$(4^{x} + 3)(4^{x} - 2) = 0$

Are you seeing it?

anonymous · Answer

yes..

tkhunny · Answer

Always look for Quadratic Forms.

$Something^{2} + a\cdot Something + b = 0$

This is why you did WAY MORE factoring problems than you thought was necessary back in Algebra.

Good work.

anonymous · Answer

but i have to write it in terms of Ln, thats what i dont understand

anonymous · Answer

$$
w=4^x\implies \ln w= x\ln4\implies x=\frac{\ln w}{\ln 4} 
$$

hartnn · Answer

did that help you further ?
which step are u still stuck on ?

anonymous · Answer

The correct answer is ln2/ln4 
how is that?

hartnn · Answer

didn't you get w =2 ??

and w=-3 is dicarded, because log og negative number is not defined.

hartnn · Answer

log of**

hartnn · Answer

$w=4^x\implies \ln w= x\ln4\implies x=\frac{\ln w}{\ln 4} =\frac{\ln 2}{\ln 4} $

anonymous · Answer

ok that makes sense now