Question

Logarithm and Indices Question!

Answer 1

Solve for x, correct to two significant figure:

\[4^{2X+1} * 5^{X-2} =6 ^{1-X}\]

I think I'll need to use a logarithm in order to sort it out, but I;m confused as to how to start, does anyone have a few hints?

I'm more interested in the method than the answer. c;

Thanks in advance!

Answer 2

\[4^{2x+1}=\frac{ 6^{1-x} \times 5}{ 5^{x-1} }=>4^{2x+1}=6^{1-x}\times 5^{1-x} \times 5\] can u do it from here :)

Answer 3

I tried it again but I'm still stuck, sorry. =/

What is the next step?

Answer 4

Wait, so I have to divide \[4^{2x+1}\]by \[4^{3x}\]

And then divide the right hand side by the same number and /then/ deal with the logarithms?

Answer 5

hey forget that^ thing cause it ws getting too messy :/ and do it like this-
\[\log_{2}(4^{2x+1}\times5^{x-2})=\log_{2}6^{1-x}\]\[\log_{2}4^{2x+1} +\log_{2}5^{x-2}=(1-x)\log_{2}2 +(1-x)\log_{2}3\]\[4x+2+(x-2)\log_{2}5=1-x+\log_{2}3-xlog_{2}3\]\[4x+2+xlog_{2}5-2\log_{2}5=1-x+\log_{2}3-xlog_{2}3\]\[x(4+\log_{2}5+1+\log_{2}3)=2\log_{2}5-2+1+\log_{2}3\]

Answer 6

So:\[4^{1-x} * 4^{3x} = 6^{1-x} * 5^{1-x} * 5\]

\[\frac{ 4^{1-x} }{ 6^{1-x } * 5^{1-x } } = \frac{ 5 }{ 4^{3x} }\]

Answer 7

Alright, this is making a bit more sense, so you actually multiplied one of the logs on the right side by (1-x) in order to "detach" it? Neat trick.

Give me a few minutes, I think I should be able to get it but let me make doubly sure. c:

Answer 8

So

\[\frac{ \log_{2}75 - 1 }{ \log_{2} 15 + 5 } = x\] = 0.587 = 0.59

Yes! Correct answer, thanks for your time and trouble @imqwerty ! ;U;

Answer 9

(: