Question

Opinions please! Suppose you were given the equation 4^2x+3=128 to solve, and there were no restrictions placed on how you solved it. Would you use algebra or your graphing calculator? Write a paragraph explaining your reasoning for your answer.

Answer 1

Algebra.

Answer 2

2x+3 is all raised

Answer 3

\[4^{2x+3}=128\]
i would take natural log of both sides

Answer 4

\[\ln(4^{2x+3})=\ln128\]

so that I could bring down the 2x+3

\[(2x+3)\ln4=\ln128\]

Answer 5

\[4 = 2^2\]\[\implies 4^{2x+3} = (2^2)^{2x+3}=2^{4x + 6}\]
\[128 = 2^7 \]
\[4^{2x+3}=128\]\[\implies 2^{4x+6} = 2^7\]\[\implies 4x + 6 = 7\]\[\implies 4x = 1\]\[\implies x = \frac{1}{4}\]

Answer 6

then i would divide both sides by ln4

\[2x+3=\frac{\ln128}{\ln4}\]

Answer 7

then i would subtract 3 on both sides

\[2x=\frac{\ln128}{\ln4}-3\]

then i would multiply both sides by 1/2

\[x=\frac{1}{2}*(\frac{\ln128}{\ln4}-3)\]

Answer 8

Two things with the same base and + between the powers can be seperated

4^2x+3 = 128
4^2x x 4^3 = 128
4^2x = 2

Now you need to use logarithims, I'm not too familiar with them. I will do my best though

log4^2x = log2

I'm pretty sure the handy thing with logarithims is it allows you to bring the power to the front of the equation. Don't quote me. I'm not entirely sure, and don't get mad if I'm wrong.

2x*log4 = log2

I'm not sure if this is right, but I know there is a method to solve the equation. Try a youtube guide on solving logarithims with exponents.

Answer 9

then i would simplify if possible

Answer 10

I like my way better, no calculator ;p

Answer 11

Thanks guys!

Answer 12

\[x=\frac{1}{2}*(\frac{\ln2^7}{\ln2^2}-3)=\frac{1}{2}*(\frac{7\ln2}{2\ln2}-3)=\frac{1}{2}(\frac{7}{2}-3)\]
\[=\frac{7}{4}-\frac{3}{2}=\frac{7-6}{4}=\frac{1}{4}\]
no calculator needed

Answer 13

lol but i do like polpak's way better because there are less steps