SOLUTION: Solve for x in each: A) 9^(2x+3) =81^(2x-1) B)2log2x=log(7x-3) C)7*3^(x+2)=5x d)log3(x+1)+log3(x+6)=2 E) log10^e

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve for x in each: A) 9^(2x+3) =81^(2x-1) B)2log2x=log(7x-3) C)7*3^(x+2)=5x d)log3(x+1)+log3(x+6)=2 E) log10^e      Log On


   

Question 1072415: Solve for x in each:
A) 9^(2x+3) =81^(2x-1)
B)2log2x=log(7x-3)
C)7*3^(x+2)=5x
d)log3(x+1)+log3(x+6)=2
E) log10^e
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
A) 9^(2x+3) =81^(2x-1)
Write 81 as 92

9%5E%282x%2B3%29=%289%5E2%29%5E%282x-1%29

Multiply the exponents on the right
to rmove the parentheses:

9%5E%282x%2B3%29=9%5E%282%282x-1%29%29

Since the exponents on each side have
the same positive base other than 1,
The exponents are equal, so

2x%2B3=2%282x-1%29

You finish.

B)2log(2x)=log(7x-3)
2%2Alog%28%282x%29%29=log%28%287x-3%29%29

Use the rule of logarithms that allows us
to move the 2 coefficent on the left up
above the (2x) as an exponent:

log%28%282x%29%5E2%29=log%28%287x-3%29%29

Now what the log is of on the left must be
equal to what the log is of on the right,
so

%282x%29%5E2=7x-3

You finish that.  It's a quadratic. 
Check both answers

C)7*3^(x+2)=5x
There is no way to solve an equation if the variable
appears both as part of exponent and also part of
a non-exponent, except as an approximation using a
graphing calculator.  However we can show with a
calculator that it has no solution because the left
side is always greater than the right side.

d)log3(x+1)+log3(x+6)=2
I can't tell whether the 3's are bases of the logarithms
or coefficients of the expressions in parentheses.

E) log10^e
"log" with no base written means that the base is 10.  So
log%2810%2C%2810%5Ee%29%29 asks "What exponent of the base is
required to give what the log is of?"  What the log is of
is 10e, and so the exponent of the base 10 that
is required to give what the log is of, namely 10e is
simply the exponent e.

Edwin