I'm stuck on this problem

[abel muroi]

I need some help on this problem...10^1-x= 6^x and was told to solve for x

Here is my work...

(1-x)log 10 = (x) log 6

1-x = 0.77...x

 

[Deleted member 4993]

I need some help on this problem...10^1-x= 6^x and was told to solve for x

Here is my work...

(1-x)log 10 = (x) log 6

1-x = 0.77...x

10^1-x= 6^x

(1-x)log 10 = (x) log 6

1 - x = 0.77815 * x

A similar problem would be

A + B*x = C*x

A = x * (C - B)

x = A/(C-B)

Follow the same procedure.....

 

[abel muroi]

10^1-x= 6^x

(1-x)log 10 = (x) log 6

1 - x = 0.77815 * x

A similar problem would be

A + B*x = C*x

A = x * (C - B)

x = A/(C-B)

Follow the same procedure.....

ok so after I get to 1= 1.778x.. What should I do next? Should I divide both sides by x?

i keep dividing but I still get the wrong answer

 

[soroban]

Hello, abel muroi!

Advice: Do not insert decimal values until the very end.

Solve for \(\displaystyle x\!:\; 10^{1-x} \;=\; 6^x\)

Take logs, base 10:

\(\displaystyle \qquad\begin{array}{c}\log(10^{1-x}) \;=\;\log(6^x) \\ \\

(1-x)\log10 \;=\; x\log 6 \\ \\

1-x \;=\;x\log6 \\ \\

x + x\log6 \;=\; 1 \\ \\

x(1+\log6) \;=\; 1 \\ \\

x \;=\;\dfrac{1}{1+\log6} \end{array}\)


Now reach for your calculator: \(\displaystyle \;x \:=\:0.562381856...\)

 

[Steven G]

10^(1-x) = 6^x ; my way:

10^1 * (10^(-x) = 6^x

6^x / 10^(-x) = 10

6^x * 10^x = 10

60^x = 10

x = LOG(10) / LOG(60) = .56238.....

Come on Denis, your 2nd line has unbalanced parenthesis. I felt that it would be better if I pointed this out to you....

 

[daon2]

Thanks JoeMoe; is my "spacing" ok

5 periods after periods after the .56238 does seem a bit excessive, if not grammatically incorrect!