SOLUTION: Determine the solutions of the following equations. Leave your answers in exact form. a.) 2^(2/log(base5)x) =1/16 b.) logbase4(x+2)=logbase4 x+logbase4 2 c.)5^2^x=3 I

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Determine the solutions of the following equations. Leave your answers in exact form. a.) 2^(2/log(base5)x) =1/16 b.) logbase4(x+2)=logbase4 x+logbase4 2 c.)5^2^x=3 I      Log On


   

Question 1002181: Determine the solutions of the following equations. Leave your answers in exact form.
a.) 2^(2/log(base5)x) =1/16
b.) logbase4(x+2)=logbase4 x+logbase4 2
c.)5^2^x=3

I tried a and got 5^64. If that is correct then I will only need help with b and c. B I know that because the bases are all 4 it makes it easier but I just don't know what steps to take. C is confusing to me because 2 is raised but the x is also raised to so that really confuses me. Thank you all in advance.
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
a).
2%5E%282%2Flog%285%2Cx%29%29+=1%2F16+
2%5E%282%2Flog%285%2Cx%29%29+=1%2F2%5E4+
2%5E%282%2Flog%285%2Cx%29%29+=2%5E%28-4+%29.........if bases same, exponents are same to
2%2Flog%285%2Cx%29+=+-4+
2=+-4log%285%2Cx%29
2%2F-4=+log%285%2Cx%29
-1%2F2=+log%285%2Cx%29 ......change the base to base 10
-1%2F2=+log%28x%29+%2Flog%285%29
-%281%2F2%29log%285%29=+log%28x%29+
log%285%5E%28-1%2F2%29%29=+log%28x%29+
log%281%2F5%5E%281%2F2%29%29=+log%28x%29+
log%281%2Fsqrt%285%29%29=+log%28x%29 .....if log same, then
x=1%2Fsqrt%285%29


b.)
log%284%2C%28x%2B2%29%29=log%284%2C+x%29%2Blog%284%2C+2%29
log%284%2C%28x%2B2%29%29=log%284%2C+2x%29...if base same and log same, we have
x%2B2=2x
2=2x-x
x=2


c.)
5%5E%282%5Ex%29=3....take the log of both sides
log%285%5E%282%5Ex%29%29=log%283%29
%282%5Ex%29log%285%29=log%283%29
2%5Ex=log%283%29%2Flog%285%29....take the log of both sides again
log%282%5Ex%29=log%28%28log%283%29%2Flog%285%29%29%29
xlog%282%29=log%28%28log%283%29%29%29-log%28%28log%285%29%29%29
xlog%282%29=log%28%28log%283%29%29%29-log%28%28log%285%29%29%29
x=%28log%28%28log%283%29%29%29-log%28%28log%285%29%29%29%29%2Flog%282%29












Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Determine the solutions of the following equations. Leave your answers in exact form.
a.) 2^(2/log(base5)x) =1/16
b.) logbase4(x+2)=logbase4 x+logbase4 2
c.)5^2^x=3

I tried a and got 5^64. If that is correct then I will only need help with b and c. B I know that because the bases are all 4 it makes it easier but I just don't know what steps to take. C is confusing to me because 2 is raised but the x is also raised to so that really confuses me. Thank you all in advance.
Sorry, but 5%5E64 is far from correct



22/(log5 x) = 1%2F16

22/(log5 x) = 2%5E-4

2%2Flog+%285%2C+x%29+=+-+4 --------- If bases are equal, their exponents are too

-+4+%2A+log+%285%2C+x%29+=+2 ------- Cross-multiplying

log+%285%2C+%28x%5E%28-+4%29%29%29+=+2

x%5E%28-+4%29+=+5%5E2 ---------- Converting to EXPONENTIAL form

x%5E%28-+4%29+=+25

highlight_green%28x+=+25%5E%28-+1%2F4%29%29, or highlight_green%28x+=+%281%2F25%29%5E%281%2F4%29%29, or 0.447213595