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Consider the equation 
10^((2z)/(3))=15.
Solve the equation for 
z. Express the solution as a logarithm in base10.

z=
Approximate the value of 
z. Round your answer to the nearest thousandth.

z~~

Consider the equation 102z3=15 10^{\frac{2 z}{3}}=15 .\newlineSolve the equation for z z . Express the solution as a logarithm in base1010.\newlinez= z= \newlineApproximate the value of z z . Round your answer to the nearest thousandth.\newlinez z \approx

Full solution

Q. Consider the equation 102z3=15 10^{\frac{2 z}{3}}=15 .\newlineSolve the equation for z z . Express the solution as a logarithm in base1010.\newlinez= z= \newlineApproximate the value of z z . Round your answer to the nearest thousandth.\newlinez z \approx
  1. Write Equation: Write down the given equation.\newlineWe have the equation 10(2z3)=1510^{\left(\frac{2z}{3}\right)} = 15.
  2. Take Logarithm: Take the logarithm of both sides of the equation to solve for zz.\newlineWe use the property that if ab=ca^b = c, then loga(c)=b\log_a(c) = b.\newlineTaking the logarithm base 1010 of both sides, we get log10(10(2z3))=log10(15)\log_{10}(10^{(\frac{2z}{3})}) = \log_{10}(15).
  3. Simplify Left Side: Simplify the left side of the equation using the property of logarithms.\newlineThe logarithm of a power is the exponent times the logarithm of the base, so log10(102z3)\log_{10}(10^{\frac{2z}{3}}) simplifies to 2z3\frac{2z}{3}.\newlineOur equation now is 2z3=log10(15)\frac{2z}{3} = \log_{10}(15).
  4. Solve for z: Solve for z.\newlineTo isolate z, we multiply both sides of the equation by 32\frac{3}{2}.\newlinez=(32)log10(15)z = \left(\frac{3}{2}\right) \cdot \log_{10}(15).
  5. Calculate z: Calculate the value of z using a calculator.\newlineUsing a calculator, we find that log10(15)\log_{10}(15) is approximately 1.1761.176.\newlineSo, z(32)×1.176z \approx \left(\frac{3}{2}\right) \times 1.176.
  6. Perform Multiplication: Perform the multiplication to find the approximate value of zz.\newlinez(32)×1.1761.764z \approx \left(\frac{3}{2}\right) \times 1.176 \approx 1.764.

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