find the solution to the following equation

Question

a)log₂(log₃x)=4

b)ln(x-1)+ln(+2)=1

 

Answer 1

a.  log₂(log₃x) = 4

 

When:
b^y = x
Then the base b logarithm of a number x:
log_bx = y

 

In this case, then,

 

2⁴ = 16 = log₃(x)

 

∴ x = 3¹⁶ = 43046721

 

b)  ln(x-1)+ln(+2)=1  

 

e^{ln(x-1)+ln(2)} = e

 

2(x - 1) = e

 

2x = e + 2

 

x = (e/2) + 1

 

 

 

 

 

x

 

 

Answer 2

a) log2[log3(x)]=4

    2^4=log3(x)

       16=log3(x)

       3^16=x

       x=43,046,721

 

b) ln(x-1)+ln(2)=1

    ln[(x-1)(2)]=1

    2(x-1)=e

    2x-2=e

    2x=e+2

    2x=2.7183+2

    2x=4.7183

      x=4.7183/2

      x=2.35915