SOLUTION: Consider the expression, {{{ r^s = A(t-p)^x }}} . An equivalent logarithmic expression is. r is the base A. {{{ s = A log ( r, (t-p))^x }}} B. {{{ s = log ( r, Ax (t-p)) }}}

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Consider the expression, {{{ r^s = A(t-p)^x }}} . An equivalent logarithmic expression is. r is the base A. {{{ s = A log ( r, (t-p))^x }}} B. {{{ s = log ( r, Ax (t-p)) }}}       Log On


   

Question 944780: Consider the expression, +r%5Es+=+A%28t-p%29%5Ex+ . An equivalent logarithmic expression is. r is the base
A. +s+=+A+log+%28+r%2C+%28t-p%29%29%5Ex+
B. +s+=+log+%28+r%2C+Ax+%28t-p%29%29+
C. +s+=+Ax+log+%28+r%2C+%28t-p%29%29+
D. +s+=+log+%28+r%2C+A%29%28t-p%29%5Ex+
Struggling with this question, can someone show me the steps used to solve this question.
Thank you
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Consider the expression, +r%5Es+=+A%28t-p%29%5Ex+ . An equivalent logarithmic expression is. r is the base
base = r
exponent = log = s
result = A(t-p)^x
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Answer: s = logr[A(t-p)^x]
Cheers,
Stan H.
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A. +s+=+A+log+%28+r%2C+%28t-p%29%29%5Ex+
B. +s+=+log+%28+r%2C+Ax+%28t-p%29%29+
C. +s+=+Ax+log+%28+r%2C+%28t-p%29%29+
D. +s+=+log+%28+r%2C+A%29%28t-p%29%5Ex+

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

X+=+B%5EY is equivalent to Y+=+log%28B%2C+X%29
you are given +r%5Es+=+A%28t-p%29%5Ex+
compare it to X+=+B%5EY
if r is base then Y=s and X++=A%28t-p%29%5Ex and Y+=+log%28B%2C+X%29 will be
s=log%28r%2CA%28t-p%29%5Ex%29%29
so, answer is D. +s+=+log+%28+r%2C+A%28t-p%29%5Ex+%29