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Question
hey. i was wondering if you could help me solve this problem. it says:
solve for x if (1/125) = 25^(9x+5)
so far i canged both sides to equal bases so i have:
5^-3 = (5^2)^(9x+5), but then i got stuck. thanxs
Questioner: astrid
Category: Advanced Math
Private: No
Subject: pre calc (exponential and logarithmic functions)
Question: hey. i was wondering if you could help me solve this problem. it says:
solve for x if (1/125) = 25^(9x+5)
so far i canged both sides to equal bases so i have:
5^-3 = (5^2)^(9x+5), but then i got stuck. thanxs
..................................
Hi, Astrid,
If:
5^-3 = (5^2)^(9x+5)
then use more of the power laws:
To raise a power to a power, keep the base, multiply exponents:
5^-3 = 5^[2(9x+5)]
5^-3 = 5^[18x+10]
Now equate the exponents:
-3 = 18x+10
You can take it from there.
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Paul Klarreich
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I can answer questions in basic to advanced algebra (theory of equations, complex numbers), precalculus (functions, graphs, exponential, logarithmic, and trigonometric functions and identities), basic probability, and finite mathematics, including mathematical induction. I can also try (but not guarantee) to answer questions on Abstract Algebra -- groups, rings, etc. and Analysis -- sequences, limits, continuity. I won't understand specialized engineering or business jargon.
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I taught at a two-year college for 25 years, including all subjects from algebra to third-semester calculus.
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