# how do I solve log7 0.915

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Evaluate `log_7 0.915` :

Use the change of base formula :`log_b c=(lnc)/(lnb)=(logc)/(logb)`

Then `log_7 0.915=(ln 0.915)/(ln7)`

These can be evaluated with a scientific calculator:

`log_7 0.915~~-0.04565`

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To 4 decimal places `log_7 0.915~~-0.0457`

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Check: `7^(-0.0457)~~0.914911359`

You should turn the given logarithm into two logarithms of base 10, using the following formula, such that:

`log_a b = (lg b)/(lg a)`

Reasoning by analogy yields:

`log_7 0.915 = (lg 0.915)/(lg 7)`

You need to evaluate `lg 0.915` such that:

`lg 0.915 = lg (915/1000) `

You need to convert the logarithm of quotient into a difference of two logarithms such that:

`lg 0.915 = lg 915 - lg 1000`

`lg 0.915 = lg 915 - lg 10^3`

Using the power property of logarithms, `log a^b = b*log a` , yields:

`lg 0.915 = lg 915 - 3 lg 10`

Since `lg 10 = 1` yields:

`lg 0.915 = lg 915 - 3`

You need to evaluate `lg 915` using the following algorithm, such that:

`100 < 915 < 1000 => lg 100 < lg 915 < lg 1000 => lg 10^2 < lg 915 < lg 10^3`

`2 < lg 915 < 3`

You need to use the following Euler's identity, such that:

`log (sqrt(x*y)) = (log x + log y)/2`

Since `100 < 915 < 1000` yields:

`lg sqrt(100*1000) = (lg 100 + lg 1000)/2 = (2 + 3)/2 = 2.5`

Since `sqrt(100*1000) < 915 < 1000` yields `2.5 < lg 915 < 3`

You can continue to calculate `sqrt(100*1000) = 316.2277` and to use Euler's identity again, such that:

`lg sqrt(316.2277*1000) = (2.5 + 3)/2 = 2.75`

Since `316.2277 < 915 < 1000` , you can continue to tighten the interval, such that:

` lg sqrt(562.3412 *1000) = (2.75+3)/2 = 2.875`

Since , you can continue to tighten the interval, such that:

`lg sqrt (749.8941*1000) = (2.875+3)/2 = 2.9375`

`sqrt (749.8941*1000) = 865.9642 < 915 < 1000`

`lg sqrt (749.8941*1000) = (2.9375 + 3)/2 = 2.96875`

Since `lg(865.9642*1000) = 930 > 915` , the process can stop.

Hence, evaluating `lg 915` yields `lg 915 ~~ 2.96875`

`lg 0.915 =2.96875 - 3 = -0.03125`

You can evaluate `(lg 7) ` using the same algorithm, such that:

`(lg 7) = 0.84509`

`log_7 0.915 = -0.03125/0.84509 = -0.0369783`

**Hence, evaluating the given logarithm, using Euler's algorithm, yields **`log_7 0.915 = -0.0369783.`