## Jadedry one year ago Logarithm and Indices Question!

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!

2. imqwerty

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

I tried it again but I'm still stuck, sorry. =/ What is the next step?

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?

5. imqwerty

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$

So:$4^{1-x} * 4^{3x} = 6^{1-x} * 5^{1-x} * 5$ $\frac{ 4^{1-x} }{ 6^{1-x } * 5^{1-x } } = \frac{ 5 }{ 4^{3x} }$

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:

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;

9. imqwerty

(: