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Logbase5(2)-logbase5(x+7) =1

Logbase5(2)-logbase5(x+7) =1-example-1

1 Answer

4 votes
Solve for x over the real numbers:
log(5, 2) - log(5, x + 7) = 1

log(5, 2) - log(5, x + 7) = log(2)/log(5) - log(x + 7)/log(5):
log(2)/log(5) - log(x + 7)/log(5) = 1

Rewrite the left hand side by combining fractions. log(2)/log(5) - log(x + 7)/log(5) = (log(2) - log(x + 7))/log(5):
(log(2) - log(x + 7))/log(5) = 1

Multiply both sides by log(5):
log(2) - log(x + 7) = log(5)

Subtract log(2) from both sides:
-log(x + 7) = log(5) - log(2)

Multiply both sides by -1:
log(x + 7) = log(2) - log(5)

log(2) - log(5) = log(2) + log(1/5) = log(2) + log(1/5) = log(2/5) = log(2/5):
log(x + 7) = log(2/5)

Cancel logarithms by taking exp of both sides:
x + 7 = 2/5

Subtract 7 from both sides:

Answer: x = -33/5
User Krish Kr
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