Logs and indices questions: $4^{2x}-2^{x+1}=48$ and $6^{2x+1}-17*{6^x}+12=0$

Question

Hi can anyone solve these two questions using logs and indices

a. $$4^{2x}-2^{x+1}=48$$

b. $$6^{2x+1}-17*{6^x}+12=0$$

Thanks.

Answer 1

I believe the last question to be $6^{2x+1}-17(6^x)+12=0$

$$\iff6(6^x)^2-17(6^x)+12=0$$

$$6^x=\frac{17\pm\sqrt{17^2-4\cdot6\cdot12}}{2\cdot6}=\frac{17\pm1}{12}=\frac32,\frac43$$

Answer 2

I am afraid but I do not think that there is any explicit solution for $$4^{2x}-2^{x+1}=48$$ To visualize it better, since the terms grow very fast, it is better to look at function $$f(x)=\log(4^{2x})-\log(2^{x+1}+48)=2x\log(4)-\log(2^{x+1}+48)$$ which is basically a straight line.

To solve this equation, let us use Newton, which, starting from a reasonable guess $x_0$, will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Here, we shall have $$f'(x)=2 \log (4)-\frac{2^{x+1} \log (2)}{2^{x+1}+48}$$ Since the function seems to be very close to linearity, let me be lazy and start iterations at $x_0=0$. Then the successive iterates of Newton are $1.42522$, $1.43474$ which is the solution for six significant figures.

What is amazing is that, if you start the calculations at $x_0=1$, the result of the first iteration is $$x_1=\frac{1}{51} \left(25+\frac{13 \log (13)}{\log (2)}\right)\approx 1.43345$$.

I also thought that there are some typo's in the equation but I give you this just for illustration purposes of a simple numerical solution of a quite difficult problem.

Answer 3

Thanks for the answers, after a night of thoughts, I came up with the idea of using quadratic equations and logarithms to solve the answers

Let 2x=a

a^2-2a-48=0

a=8 or a=-6(rejected)

2^x=3

log(2)8=3

x=3

Let 6x=a

6a^2-17a+12=0

a=1.5 or4/3

log(6)1.5=0.226

log(6)1.5=0.1606

x=0.226 or 0.1606

Cheers!