SOLUTION: Solve for: a){{{ (squareroot of 2 ^ (squareroot of x)) = (squareroot of 2^x) }}} b) {{{ 2 ^x^2 = (2^x)^2}}} c) {{{ 2 ^(x+3) = 2 ^x +3}}} d) {{{ log(3x) = 3 log x }}} Th

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Solve for: a){{{ (squareroot of 2 ^ (squareroot of x)) = (squareroot of 2^x) }}} b) {{{ 2 ^x^2 = (2^x)^2}}} c) {{{ 2 ^(x+3) = 2 ^x +3}}} d) {{{ log(3x) = 3 log x }}} Th      Log On


   

Question 120721: Solve for:
a)+%28squareroot+of+2+%5E+%28squareroot+of+x%29%29+=+%28squareroot+of+2%5Ex%29++
b) +2+%5Ex%5E2+=+%282%5Ex%29%5E2
c) +2+%5E%28x%2B3%29+=+2+%5Ex+%2B3
d) +log%283x%29+=+3+log+x+

Thank you so much! :)
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do b, c, and d. Your rendering for part a is ambiguous. Do you mean sqrt%282%29%5Esqrt%28x%29=sqrt%282%5Ex%29 or sqrt%282%5Esqrt%28x%29%29=sqrt%282%5Ex%29?

b.) 2+%5Ex%5E2+=+%282%5Ex%29%5E2

Take the base-2 log of both sides:

log%282%2C2%5Ex%5E2%29=log%282%2C%282%5Ex%29%5E2%29

Remembering that log%28b%2Cx%5En%29=n%2Alog%28b%2Cx%29 we can say:

x%5E2%2Alog%282%2C2%29=2%2Alog%282%2C%282%5Ex%29%29 and then apply the rule again to say:

x%5E2%2Alog%282%2C2%29=2x%2Alog%282%2C%282%29%29

Now remember that log%28b%2Cb%29=1 we can simplify to:

x%5E2=2x, or

x=2

c. 2+%5E%28x%2B3%29+=+2+%5Ex+%2B3

Remember that b%5E%28a%2Bc%29=b%5Ea%2Ab%5Ec, so:

2%5Ex%2A2%5E3=2%5Ex%2B3

Add -2%5Ex to both sides:

2%5Ex%2A2%5E3-2%5Ex=3

Factor out 2%5Ex

%282%5Ex%29%282%5E3-1%29=3
%282%5Ex%29%288-1%29=3
%282%5Ex%29%287%29=3
2%5Ex=3%2F7

Take the log base 2 of both sides:

log%282%2C2%5Ex%29=log%282%2C3%2F7%29

Again, using log%28b%2Cx%5En%29=n%2Alog%28b%2Cx%29 we can say:

x%2Alog%282%2C2%29=log%282%2C3%2F7%29

x=log%282%2C3%2F7%29 or x=log%282%2C3%29-log%282%2C7%29 if you prefer.




d. log%283x%29+=+3log%28x%29

Using log%28b%2Cx%5En%29=n%2Alog%28b%2Cx%29 in reverse we can say:

log%283x%29+=+log%28x%5E3%29

Take the antilog of both sides:

3x=x%5E3

x%5E2=3

x=sqrt%283%29 Only the positive square root will do because the domain of the log%28b%2Cx%29 function is constrained to 0%3Cx%3C%2Binfinity