Asked by Ian
solve and graph: log4(x+5)=3. Give me the algebraically and graphically answers
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Algebraic solution
log_4(x + 5) = 3
=> x + 5 = 4^3 (convert to exponential form)
=> x + 5 = 64
=> x = 64 − 5 = 59.
Check domain: x + 5 > 0 ⇒ x > −5. 59 is valid. So the solution is x = 59 (point of intersection (59, 3)).
Graphical explanation
Plot the two curves:
- y = log_4(x + 5) — a logarithm with vertical asymptote at x = −5, shifted left 5 units from y = log_4(x). It is increasing.
- y = 3 — a horizontal line.
Key points on y = log_4(x + 5):
- x = −4 → y = log_4(1) = 0 (point (−4, 0))
- x = −1 → y = log_4(4) = 1 (point (−1, 1))
- x = 11 → y = log_4(16) = 2 (point (11, 2))
- x = 59 → y = log_4(64) = 3 (point (59, 3))
The horizontal line y = 3 intersects the log curve at (59, 3), so x = 59 is the solution.
log_4(x + 5) = 3
=> x + 5 = 4^3 (convert to exponential form)
=> x + 5 = 64
=> x = 64 − 5 = 59.
Check domain: x + 5 > 0 ⇒ x > −5. 59 is valid. So the solution is x = 59 (point of intersection (59, 3)).
Graphical explanation
Plot the two curves:
- y = log_4(x + 5) — a logarithm with vertical asymptote at x = −5, shifted left 5 units from y = log_4(x). It is increasing.
- y = 3 — a horizontal line.
Key points on y = log_4(x + 5):
- x = −4 → y = log_4(1) = 0 (point (−4, 0))
- x = −1 → y = log_4(4) = 1 (point (−1, 1))
- x = 11 → y = log_4(16) = 2 (point (11, 2))
- x = 59 → y = log_4(64) = 3 (point (59, 3))
The horizontal line y = 3 intersects the log curve at (59, 3), so x = 59 is the solution.
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