You don't need logarithms.
.
Now make the substitution suggested and solve the resulting quadratic.
Hi guys, I am unsure how this book has come about with the answer to this question as I thought I had it worked out.
Question:
Use the substitution y = 2^x to solve the following equation giving your answer in exact form involving logarithms to the base ten.
2^(2x) - 2^(x+3) + 15 = 0
My work goes like this:
y^2 - 2^(x+3) + 15 = 0
y^2 + 15 = 2^(x+3) <- Expand the log
log y^2 + log 15 = log 2^(x+3)
= (x+3)log 2
= xlog 2 + 3log2
= log 2^x + log 2^3
= log y + log 8 <- Rearrange the equation
log y^2 - log y = log 8 - log 15
log ((y^2)/y) = (log 8/15)
y = 8/15
The answers however state the following and I am unsure how they come about as I thought my working was sound.
log 3 / log 2 & log 5 / log 2
Any help would be greatly appreciated.
Cheers
Ben
But you didn't make the substitution completely!
Your equation should be
which is a simple quadratic equation.
and this is also wrong: log(a+ b) is NOT equal to log(a)+ log(b).y^2 + 15 = 2^(x+3) <- Expand the log
log y^2 + log 15 = log 2^(x+3)
so NOW you have finished the substitution- but the left side, log(y^2)+ log(15), is still wrong.= (x+3)log 2
= xlog 2 + 3log2
= log 2^x + log 2^3
= log y + log 8 <- Rearrange the equation
log y^2 - log y = log 8 - log 15
log ((y^2)/y) = (log 8/15)
y = 8/15
The answers however state the following and I am unsure how they come about as I thought my working was sound.
log 3 / log 2 & log 5 / log 2
Any help would be greatly appreciated.
Cheers
Ben