Supposing that you want to solve the expression `(yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy), ` you need to take logarithm of this expression such that
`log((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = log((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy))` Converting the logarithm of the product into a sum of logarithms yields:
`log((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = log((yz)^(log y-log z)) + log ((zx)^(log z-logx))+ log ((xy)^(log x -logy))` Using the logarithmic identities yields:
`log((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = (log y-log z)(log (yz)) + (log z-logx)(log(xz)) + (log x -logy)(log(xy))`
You need to convert the logarithm of product into the sum of logarithms such that:
`log((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = (log y - log z)(log y + log z) + (log z - log x)(log z+ log x) + (log x - log y)(log x +log y)`
You should convert the special products into differences of squares such that:
`log((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = log^2y - log^2 z + log^2 z - log^2 x + log^2 x - log^2 y`
Reducing like terms yields:
`log((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = 0 =gt ((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = 10^0`
`((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = 1`
Hence, evaluating the value of given expression yields `((yz)^(log y-log z)*(zx)^(log z-logx)*(xy)^(log x -logy)) = 1.`
We’ll help your grades soar
Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.
- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support
Already a member? Log in here.
Are you a teacher? Sign up now