Question

Solve for x:
log base 64 25x-log base 64 (x-7)=2/3

Answer 1

The equation is


`\log_(64)(25x)-\log_(64)(x-7)=(2)/(3)`.

First, from the rule
`\log_ba-\log_bc=\log_b{(a)/(c)}`, we rewrite the left hand side as:


`\displaystyle{ \log_(64){ (25x)/(x-7) }`.

Now we write the right hand side as a log with base 64, so that we can make the arguments equal, setting a new equation.


`\displaystyle{ (2)/(3)= ((2)/(3))\log_(64)64=\log_(64)64^{(2)/(3)}`.

We can simplify the argument of this expression into a more manageable expression using the laws of exponents:


`64^{(2)/(3)}=(2^6)^{(2)/(3)}=2^{6\cdot(2)/(3)}=2^4=16`.


Thus, from
`\displaystyle{ \log_(64)({ (25x)/(x-7) })=\log_(64)(16)` we clearly have:


`\displaystyle{ (25x)/(x-7)=16`.

For x different from 7:

25x=16(x-7)

25x=16x-112

9x=-112

x=-112/9.

This value is approximately -12.44, so x-7, the argument of both the first and the second expression, would be negative. Thus, the equation has no solution.