Exponential and logarithmic equations

When does a hypothetical quantity starting at 100 and doubling every 3 hours reach 300? The equation finds time; a logarithm recovers an exponent. Use candidate checks, exponentials and logarithms.

$100\cdot2^{t/3}=300\iff2^{t/3}=3\iff t=3\log_2 3\approx4.755.$

Time t is in hours, t≥0. The decimal is approximate; substituting the exact answer gives $100\cdot2^{\log_2 3}=300$ . For a>0, a≠1, equal powers aᵘ=aᵛ imply u=v. Different bases can be handled by taking logarithms of both positive sides. Zero and negative arguments are excluded.

Reaching a threshold

Q(t)=100·2ᵗᐟ³; Q=300

Solid Q(t)=100·2ᵗᐟ³, dashed threshold 300. Crossing t=3log₂3≈4.755 h, hypothetical quantity. Uniform scales, continuous time.

A substitution has a range

$4^x-3\cdot2^x-4=0;\quad u=2^x>0;\quad(u-4)(u+1)=0.$

Discard u=−1 because an exponential is positive. u=4 gives x=2; check 16−12−4=0. Returning to x is compulsory.

Preserve the original logarithm domains

$\log_2(x-1)+\log_2(x-3)=3;\qquad x>3.$

Both original arguments must be positive. Only there may the logarithms be combined:

$(x-1)(x-3)=8\iff(x-5)(x+1)=0.$

Candidates are 5 and −1; only 5 remains. It gives log₂4+log₂2=3. A product’s logarithm may exist where the separate logarithms do not: do not enlarge the domain.

A system

$\log_2x+y=3,\quad\log_2x-y=1\quad(x>0).$

Set u=log₂x. Adding and subtracting gives u=2, y=1, hence x=4. Both original equations hold. OpenStax Return to original arguments, signs and the physical domain before reporting an answer.