Question

Log₂(3*2^(x-1)-1)=2x-1
Can't seem to reach the answer (1 and 0)

Answer 1

\[\log_{2} (3\times2^{x-1}-1)=2x-1 \]
What I tried so far:
3*2^(x-1)-1=2^(2x-1)
3*2^(x)*2^(-1)-1=2^(2x)*2^(-1)
2^(x) = t
3*t*2^(-1)-1=t^(2)*2^(-1)
t*(1/2)=t^(2)*(1/2) /: 1/2 , /: t
t=0, t=1
2^x=0, 2^x=1
no solution, 2^x=2^0
x=0

Answer 2

i hope this helps http://www.mathportal.org/calculators.php

Answer 3

\(\normalsize\color{blue}{ \log_2(3\times 2^{x-1}-1)=2x-1\LARGE\color{white}{ \rm │ }}\)

Answer 4

\(\normalsize\color{blue}{ \log_2(3\times 2^{x-1}-1)=2x-1\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ \log_2(3\times 2^{x-1}-1)=(2x-1)\log_2(2)\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ \log_2(3\times 2^{x-1}-1)=\log_2(2^{2x-1})\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ 3\times 2^{x-1}-1=2^{2x-1}\LARGE\color{white}{ \rm │ }}\)

Answer 5

\(\normalsize\color{blue}{ (3/2)2^{x}-1=(1/2)2^{x+x} \LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ (3)2^{x}-2=2^{x+x} \LARGE\color{white}{ \rm │ }}\) is what I am up to so far.

Answer 6

Pilot, please share what you got, because I am not very sure...

Answer 7

let u = 2^x and solve as a quadratic

Answer 8

3u-2=2u^2 => 2u^2 - 3u +2 = 0 => (2u-1)(u-1) => u = 1 and u = 1/2

so 2^x = 1 and 2^x = 1/2 => x = 0 and x = -1

Answer 9

but -1 can't be an answer because log can't = -. so either i goofed on factoring or solutions given are incorrect.

Answer 10

I tried inputting it into wolfram, they seem to be in order http://www.wolframalpha.com/input/?i=Log2%283*2%5E%28x-1%29-1%29%3D2x-1

Answer 11

but i just can't tell where's the mistake

Answer 12

0 can't be a solution because it leads to an false statement.

hold on 1 sec...

Answer 13

x=0 in the original equation:
log2(3*2^(0-1)-1) = 2*0-1
log2(1/2) = -1
-1 = -1

Answer 14

\[3\cdot 2^{x-1}-1=2^{2x-1} \Rightarrow \frac{3}{2}2^x-1=\frac{1}{2}2^{2x} \Rightarrow \frac{1}{2}\left(2^x \right)^2-\frac{3}{2}\left( 2^x \right)-1\]Let \(u=2^{x}\)\[\Rightarrow u^2-3u+2 = 0 \Rightarrow \left( u-2 \right)\left( u-1 \right)=0 \Rightarrow u=2 \text{ , } u=1\]\[\Rightarrow 2^x=2 \text{ and }2^x=1\Rightarrow x=1 \text{ and } x=0\]
But \(x\ne 0\) because RHS is then negative in the original equation.
So x=1 is the only solution.

Answer 15

oops, disregard the last part about the negative. the argument of log can't be negative, not the value.

Answer 16

logs are just exponents and negative exponents are perfectly okay!

Answer 17

Nice idea, but for u, 3x would be ?

Answer 18

It won't be just 3u, will it ?

Answer 19

3x? it should be 3*2^x

Answer 20

u=2^x

Answer 21

no, you got it...

Answer 22

no, the original is \[\log_{2} \left( 3\cdot 2^{x-1}-1 \right)=2x-1\]

Answer 23

all good? there is a typo... should be \[\frac{1}{2}(2^x)^2−\frac{3}{2}(2^x)+1\] not\[\frac{1}{2}(2^x)^2−\frac{3}{2}(2^x)-1\]

Answer 24

yeah, great find pilot. I'm going to use your way to solve the other similar exercises. much appriciated.

Answer 25

you're welcome! glad i could help!

Answer 26

it's a fairly common "trick" but not always obvious. in calc, it's used all the time but kinda in reverse; anytime you do the chain rule with derivatives.